百题分类

将一百题共分为22个类别

具体频率统计

脑图统计

百题题解

1. 1. 两数之和

// 哈希表
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> unMap;
        
        for(int i{};i<nums.size();++i){
            if(unMap.find(target - nums[i]) != unMap.end()){
                return {i,unMap[target - nums[i]]};
            }

            unMap[nums[i]] = i;
        }

        return {};
    }
};
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2. 2. 两数相加

// 链表+数学相加进位

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode *head{};
        ListNode *tail{};


        int carry{};
        while(l1 || l2){
            int num1 = l1 ? l1->val : 0;
            int num2 = l2 ? l2->val : 0;

            int sum = num1 + num2 + carry;

            if(!head){
                head = tail = new ListNode(sum%10);
            }else{
                tail->next = new ListNode(sum%10);;
                tail = tail->next;
            }

            carry = sum/10;

            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        }

        if(carry > 0) tail->next= new ListNode(carry);


        return head;
    }
};
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3. 3. 无重复字符的最长子串

// 双指针
class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        unordered_map<char,int>need;

        int n = s.size();
        if(0 == n) return 0;

        int left{},right{1},result{1};
        need[s[left]]++;
        for(;right < n;++right){
            if(need[s[right]] > 0){
                while(need[s[right]] > 0){
                    need[s[left++]]--;
                }
            }
            need[s[right]]++;
            result = max(result,right-left+1);
        }

        return result;
    }
};
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4. 4. 寻找两个正序数组的中位数

// 双指针

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int l1 = nums1.size();
        int l2 = nums2.size();

        vector<int> result(l1+l2);

        l1-=1;
        l2-=1;
        int n = result.size()-1;
        while(l1 >=0 || l2 >=0){
            if(l2 >=0 && l1 >=0){
                if(nums2[l2] > nums1[l1]){
                    result[n--] = nums2[l2--];
                }else{
                    result[n--] = nums1[l1--];
                }
            }else if(l1 >= 0){
                result[n--] = nums1[l1--];
            }else{
                result[n--] = nums2[l2--];
            }
            
        }

        n = result.size() - 1;

        if(n%2 == 0) return (double)result[n/2];
        else {
            long long sum = result[n/2] + result[n/2+1];
            return double(sum)/2;
        }
    }
};
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5. 5. 最长回文子串

// 动态规划

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.size();
        // dp,拆解子问题：dp(i+1,j+1)为回文子串 并且size 》 2 ，那么dp[i,j]也是回文子串
        vector<vector<int>> dp(n,vector<int>(n));

        for(int i{};i<n;++i){
            dp[i][i] = 1;
        }

        int result{},begin{};
        for(int length{2};length <= n;++length){
            for(int i{};i<n-length+1;++i){
                if(2 == length){
                    if(s[i] == s[i+length-1]){
                        dp[i][i+length-1] = 1;
                        begin = i;
                        result = 2;
                    }
                }else{
                    if(s[i] == s[i+length-1] && dp[i+1][i+length-2] == 1){
                        dp[i][i+length-1] = 1;
                        begin = i;
                        result = length;
                    }
                }
            }
        }

        return result > 1 ? s.substr(begin,result) : s.substr(0,1);
    }
};
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6. 10. 正则表达式匹配

// 动态规划

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size();
        int n = p.size();
        // 1.状态定义：dp[i][j]表示前i个字符与p中前j个字符是否能够匹配
        vector<vector<int>> dp(m+1,vector<int>(n+1));
        dp[0][0] = true;
        auto match = [&](int i,int j){
            if(i == 0) return false;
            if(p[j-1] == '.') return true;

            if(s[i-1] == p[j-1]) return true;

            return false;
        };

        for(int i{};i<=m;++i){
            for(int j{1};j<=n;++j){
                if(p[j-1] == '*'){
                    dp[i][j] |= dp[i][j-2];

                    if(match(i,j-1)){
                        dp[i][j] |= dp[i-1][j];
                    }
                }else{
                    if(match(i,j)){
                        dp[i][j] |= dp[i-1][j-1];
                    }
                }
            }
        }

        return dp[m][n];
    }
};
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7. 11. 盛最多水的容器

// 双指针
class Solution {
public:
    int maxArea(vector<int>& height) {
        int n = height.size();
        int left{};
        int right = n-1;

        int result{};
        while(left < right){
            int h = min(height[left],height[right]);
            int width = right - left;

            result = max(result,h*width);

            if(h < height[left]){
                right--;
            }else{
                left++;
            }
        }

        return result;
    }
};
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8. 15. 三数之和

// 双指针
class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;

        int n = nums.size();

        sort(nums.begin(),nums.end());

        for(int i{};i<n-2;++i){
            if(i && nums[i] == nums[i-1])continue;
            int left = i+1;
            int right = n-1;

            while(left < right){
                if(nums[i] + nums[left] + nums[right] == 0){
                    result.push_back({nums[i],nums[left],nums[right]});

                    while(left < right && nums[left] == nums[left+1]){
                        ++left;
                    }

                    while(left < right && nums[right] == nums[right-1]){
                        --right;
                    }

                    --right;
                    ++left;
                }else if(nums[i] + nums[left] + nums[right] < 0){
                    ++left;
                }else{
                    --right;
                }
            }
        }
        return result;
    }
};
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9. 17. 电话号码的字母组合

// 递归+回溯
class Solution {
public:
    vector<string> v{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    vector<string> result;
    string path;

    void recursive(string digits,int index){
        if(path.size() == digits.size()){
            result.push_back(path);
            return;
        }

        string s = v[digits[index] - '0'];

        for(int i {};i<s.size();++i){
            path.push_back(s[i]);

            recursive(digits,index+1);

            path.pop_back();
        }
    }

    vector<string> letterCombinations(string digits) {
        if(digits.size() == 0) return {};
        recursive(digits,0);

        return result;
    }
};
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11. 19. 删除链表的倒数第 N 个结点

// 链表+数学计算
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int count{};

        // 删除倒数第n个，即删除第count-n+1

        ListNode* cNode = head;

        while(cNode != nullptr){
            cNode = cNode->next;
            count++;
        }

        ListNode *cNodeX = new ListNode(0);
        cNodeX->next = head;
        count = count - n +1;

        ListNode *node = cNodeX;
        while(count--){
            if(0 == count){
                node->next = node->next->next;
                break;
            }
            node = node->next;
        }

        return cNodeX->next;
    }
};
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12. 20. 有效的括号

// 栈
class Solution {
public:
    bool isValid(string s) {
        stack<char> st;

        int n = s.size();

        for(int i{};i<n;++i){
            if(s[i] == '('){
                st.push(')');
            }else if(s[i] == '['){
                st.push(']');
            }else if(s[i] == '{'){
                st.push('}');
            }else{
                if(st.empty() || st.top() != s[i]) return false;

                st.pop();
            }
        }

        return st.empty();
    }
};
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13. 21. 合并两个有序链表

// 链表+递归
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
        if(!list1) return list2;
        if(!list2) return list1;

        ListNode* node{nullptr};

        if(list1->val < list2->val){
            node = list1;
            list1 = list1->next;
        }else{
            node = list2;
            list2 = list2->next;
        }

        node->next = mergeTwoLists(list1,list2);

        return node;
    }
};
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14. 22. 括号生成

// 递归
class Solution {
public:
    vector<string> result;
    void solve(int open,int close,string str){
        if(0 == open && 0 == close){
            result.push_back(str);
            return;
        }

        if(open > 0){
            string s = str;
            s.push_back('(');
            solve(open-1,close,s);
        }

        if(close > open){
            string s = str;
            s.push_back(')');
            solve(open,close-1,s);
        }
    }
    vector<string> generateParenthesis(int n) {
        int open = n;
        int close = n;

        solve(open,close,"");

        return result;
    }
};
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15. 23. 合并K个升序链表

// 链表

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int k = lists.size();

        vector<int> v;

        for(int i{};i<k;++i){
            ListNode* node = lists[i];

            while(node){
                v.push_back(node->val);
                node = node->next;
            }
        }

        sort(v.begin(),v.end());

        ListNode* result = new ListNode(0);
        ListNode* node = result;

        for(int i{};i<v.size();++i){
            node->next = new ListNode(v[i]);
            node = node->next;
        }

        return result->next;
    }
};
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16. 31. 下一个排列

// 哈希表+模拟
class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int n = nums.size();
        // 先找较小数
        int i = n-2;
        for(;i>=0;--i){
            if(nums[i] < nums[i+1]){
                break;
            }
        }

        // 再找较大数
        int j = n-1;
        if(i >= 0){
            for(;j>=0;--j){
                if(nums[j] > nums[i]){
                    break;
                }
            }
            swap(nums[i],nums[j]);
        }
        
        reverse(nums.begin() + i+1,nums.end());
    }
};
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17. 32. 最长有效括号

// 动态规划
class Solution {
public:
    int longestValidParentheses(string s) {
        int n = s.size();
        // 1.dp数组定义：dp[i] 表示s[i]字符结尾的的最长有效括号字串的长度
        vector<int> dp(n,0);
        int result{};
        // 3.确定便利顺序，从下标1开始遍历即可
        for(int i{1};i<n;++i){
            if('(' == s[i]){
                continue;
            }
            // 2.确定递推公式
            if(s[i-1] == '('){
                dp[i] = 2;
                if(i-2 >= 0){
                    dp[i]+=dp[i-2];
                }
            }else if(dp[i-1] > 0){
                if((i-dp[i-1]-1) >= 0 && s[i-dp[i-1]-1] == '('){
                    dp[i] = dp[i-1]+2;
                    if((i-dp[i-1]-2) >= 0){
                        dp[i] += dp[i-dp[i-1]-2];
                    }
                }
            }

            result = max(result,dp[i]);
        }

        return result;
    }
};
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18. 33. 搜索旋转排序数组

// 二分查找
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left{},right = nums.size()-1;

        while(left <= right){
            int mid = (left + right)/2;

            if(nums[mid] == target){
                return mid;
            }else if(nums[mid] >= nums[left]){
                if(target <= nums[mid] && target >= nums[left]){
                    right = mid-1;
                }else{
                    left = mid +1;
                }
            }else{
                if(target >= nums[mid] && target <= nums[right]){
                    left = mid + 1;
                }else{
                    right = mid -1;
                }
            }
        }

        return -1;
    }
};
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19. 34. 在排序数组中查找元素的第一个和最后一个位置

// 二分查找
class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        // 先找数组中第一个不小于target的元素
        auto left = lower_bound(nums.begin(),nums.end(),target);
        // 找数组中大于目标值的第一个元素
        auto right = upper_bound(nums.begin(),nums.end(),target);
        if(left == right) return {-1,-1};

        return {(int)(left-nums.begin()),(int)(right-nums.begin() -1)};
    }
};
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20. 39. 组合总和

// 递归+回溯
class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;
    void backTracking(vector<int>& candidates,int target,int sum,int index){
        if(sum > target){
            return;
        }
        if(target == sum){
            result.push_back(path);
            return;
        }

        for(int i{index};i<candidates.size();++i){
            sum+=candidates[i];
            path.push_back(candidates[i]);
            // 递归传进去的是i,不是index，也不是index+1，因为不需要进行递归层数的限制
            backTracking(candidates,target,sum,i);
            sum-=candidates[i];
            path.pop_back();
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backTracking(candidates,target,{},0);
        return result;
    }
};
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21. 42. 接雨水

// 单调栈
class Solution {
public:
    int trap(vector<int>& height) {
        int ans = 0;
        stack<int> stk;
        int n = height.size();
        for (int i = 0; i < n; ++i) {
            while (!stk.empty() && height[i] > height[stk.top()]) {
                int top = stk.top();
                stk.pop();
                if (stk.empty()) {
                    break;
                }
                int left = stk.top();
                int currWidth = i - left - 1;
                int currHeight = min(height[left], height[i]) - height[top];
                ans += currWidth * currHeight;
            }
            stk.push(i);
        }
        return ans;
    }
};

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22. 46. 全排列

// 递归+回溯
class Solution {
public:
    // 核心是要利用used数组
    vector<vector<int>> result;
    vector<int> path;
    void backTracking(vector<int>& nums,vector<bool>& used){
        if(nums.size() == path.size()){
            result.push_back(path);
            return;
        }

        for(int i{};i<nums.size();++i){
            if(used[i] == true) continue;

            used[i] = true;
            path.push_back(nums[i]);
            backTracking(nums,used);
            path.pop_back();
            used[i] = false;
        }
    }
    vector<vector<int>> permute(vector<int>& nums) {
        vector<bool> used(nums.size());
        backTracking(nums,used);
        return result;
    }
};
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23. 48. 旋转图像

// 数学逻辑
class Solution {
public: 
    // 重点是找到规律：对于矩阵中第 i 行的第 j 个元素，在旋转后，它出现在倒数第 i 列的第 j 个位置。
    // 即matrix[r][c] 在旋转后，新的位置是matrix[c][n-r-1]

    // 借助辅助数组
    // void rotate(vector>& matrix) {
    //     auto mNew = matrix;
    //     int n = matrix.size();
    //     for(int i{};i
    //         for(int j{};j
    //             mNew[j][n-i-1] = matrix[i][j];
    //         }
    //     }

    //     matrix = mNew;
    // }
    // 利用翻转代替旋转
    // 水平翻转：matrix[r][c] = matrix[n-row-1][c]
    // 主对角线翻转: matrix[r][c] = matrix[c][r]
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        
        for(int i{};i<n/2;++i){
            for(int j{};j<n;++j){
                swap(matrix[i][j] , matrix[n-i-1][j]);
            }
        }

        for(int i{};i<n;++i){
            for(int j{};j<i;++j){
                swap(matrix[i][j] , matrix[j][i]);
            }
        }
    }
};
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24. 49. 字母异位词分组

// 哈希表
class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string,vector<string>> unMap;

        for(int i{};i<strs.size();++i){
            string s = strs[i];
            sort(s.begin(),s.end());

            unMap[s].push_back(strs[i]);
        }
        vector<vector<string>> result;
        for(auto [key,value] : unMap){
            result.push_back(value);
        }

        return result;
    }
};
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25. 53. 最大子数组和

// 贪心
class Solution {
public:
    // 贪心
    int maxSubArray(vector<int>& nums) {
        int result{INT_MIN};
        int sum{};
        for(int i{};i<nums.size();++i){
            sum+=nums[i];
            if(sum > result) result = sum;
            if(sum < 0) sum = 0;
        }

        return result;
    }
};
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26. 55. 跳跃游戏

// 贪心
class Solution {
public:
    bool canJump(vector<int>& nums) {
        int n = nums.size();

        int maxJump{};
        for(int i{};i<n;++i){
            maxJump = max(maxJump,i+nums[i]);

            if(maxJump >= n-1) return true;
            if(maxJump < i+1){
                return false;
            }
        }

        return true;
    }
};
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27. 56. 合并区间

// 贪心
class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>> result;
        int n = intervals.size();

        sort(intervals.begin(),intervals.end());

        bool falg{false};
        for(int i{1};i<n;++i){
            int start = intervals[i-1][0];
            int end   = intervals[i-1][1];

            while(i<n && intervals[i][0] <= end){
                end = max(intervals[i][1],end);
                if(i == n-1) falg = true;
                ++i;
            }
            result.push_back({start,end});
        }

        if(!falg) result.push_back(intervals.back());

        return result;
    }
};
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28. 62. 不同路径

// 动态规划
class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp(m,vector<int>(n));

        for(int i{};i<m;++i) dp[i][0] = 1;
        for(int j{};j<n;++j) dp[0][j] = 1;
        

        for(int i{1};i<m;++i){
            for(int j{1};j<n;++j){
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }

        return dp[m-1][n-1];
    }
};
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28. 64. 最小路径和

// 动态规划
class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();

        vector<vector<int>> dp(m,vector<int>(n));

        dp[0][0] = grid[0][0];
        for(int i{1};i<m;++i) dp[i][0] = grid[i][0] + dp[i-1][0];
        for(int i{1};i<n;++i) dp[0][i] = grid[0][i] + dp[0][i-1];

        for(int i{1};i<m;++i){
            for(int j{1};j<n;++j){
                dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j];
            }
        }

        return dp[m-1][n-1];
    }
};
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29. 70. 爬楼梯

// 动态规划
class Solution {
public:
    int climbStairs(int n) {
        // fn = fn-1+fn-2
        if(n <= 3) return n;

        vector<int> dp(n+1);
        dp[2] = 2;
        dp[3] = 3; 
        for(int i{4};i<=n;++i){
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
};
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30. 72. 编辑距离

// 动态规划
class Solution {
public:
	// 583.两个字符串的删除操作的变形
    using LL = unsigned long long;
    int minDistance(string word1, string word2) {
        int len1 = word1.size();
        int len2 = word2.size();
		// 1.定义dp数组以及数组下标的含义
		// dp[i][j]:字符串word1[0,i] 与word2[0,j]，将 word1 转换成 word2 所使用的最少操作数为dp[i][j]
        vector<vector<LL>> dp(len1+1,vector<LL>(len2+1));
        // 3.dp数组初始化
        for(int i{};i<=len1;++i) dp[i][0] = i;
        for(int j{};j<=len2;++j) dp[0][j] = j;
        // 4.确定遍历顺序
        for(int i{1};i<=len1;++i){
            for(int j{1};j<=len2;++j){
            	// 2.确定递推公式
                if(word1[i-1] == word2[j-1]){
                    dp[i][j] = dp[i-1][j-1];
                }else{
                    dp[i][j] = min({dp[i-1][j-1],dp[i][j-1],dp[i-1][j]}) + 1;
                }
            }
        }

        return dp[len1][len2];
    }
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31. 75. 颜色分类

// 双指针求解:经典的荷兰国旗问题
class Solution {
public:
    void sortColors(vector<int>& nums) {
        int p0{},p1{};

        int n = nums.size();

        for(int i{};i<n;++i){
            if(nums[i] == 1){
                swap(nums[i],nums[p1++]);
            }else if(nums[i] == 0){
                swap(nums[i],nums[p0]);
                if(p0 < p1){
                    swap(nums[i],nums[p1]);
                }
                ++p0;
                ++p1;
            }
        }
    }
};

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32. 76. 最小覆盖子串

// 滑动窗口+哈希表
class Solution {
public:
    string minWindow(string s, string t) {
        vector<int> need(128);

        for(char c : t) need[c]++;

        int left{},right{};
        int start{};
        int count = t.size();
        int size = INT_MAX;

        for(;right < s.size();++right){
            char c = s[right];
            if(need[c] > 0) count--;// 进行一次匹配
            need[c]--;// 将右边的字符+入窗口内

            if(count == 0){ // 当都匹配到以后才需要进行窗口的缩减
                // 缩减窗口（因为题目要求是最小子串，所以需要让need[s[left]] == 0）
                while(left < right && need[s[left]] < 0){
                    need[s[left++]]++;
                }

                if(right - left + 1 < size){    
                    size = right - left + 1;
                    start = left;
                }

                need[s[left++]]++;// 左边界右移
                count++;// 哈希表又存储了一次，所以count要++
            }
        }

        return size == INT_MAX ? "" : s.substr(start,size);
    }
};
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33. 78. 子集

// 回溯
class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;

    void recursive(vector<int>& nums,int index){
        if(index > nums.size()){
            return;
        }

        result.push_back(path);

        for(int i = index;i<nums.size();++i){
            path.push_back(nums[i]);
            recursive(nums, i+1);
            path.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        recursive(nums,0);
        return result;
    }
};
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34. 79. 单词搜索

// 递归+回溯（网格类型）
class Solution {
public:
    //定义方向
    int dx[4]{1,0,-1,0};
    int dy[4]{0,-1,0,1};
    
    //递归+回溯
    bool dfs(vector<vector<char>>& board,string word,int k,int i,int j,vector<vector<int>> &visit){
    	// 递归终止条件
        if(word[k] != board[i][j]){
            return false;
        }else if(k == word.size() -1){
            return true;
        }

        visit[i][j] = 1; // 表示已经被访问过
        bool result = false;
        // 横向遍历四个方向
        for(int d{};d<4;++d){
        	// 设置方向
            int i1 = i + dx[d];
            int j1 = j + dy[d];

            if(min(i1,j1) < 0 || i1 >=board.size() || j1 >= board[0].size()){
                continue;
            }

            if(1 == visit[i1][j1]) continue;
            // 根据方向纵向遍历
            if(dfs(board,word,k+1,i1,j1,visit)){
                result = true;
                break;
            } 
        }
        
        visit[i][j] = 0; // 回溯
        return result;
    }
    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size(),n = board[0].size();
        vector<vector<int>> visit(m,vector<int>(n));
        // 先暴力，以每个字符为起点进行递归
        for(int i{};i<m;++i){
            for(int j{};j<n;++j){
                bool flag{false};
                flag = dfs(board,word,0,i,j,visit);
                if(flag) return true;
            }
        }
        return false;
    }
};
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35. 84. 柱状图中最大的矩形

// 单调栈
class Solution {
public:
    int largestRectangleArea(vector<int>& heights)
	{
	    int ans = 0;
	    vector<int> st;
	    // 增加一个哨兵、与链表的哨兵节点一致
	    heights.insert(heights.begin(), 0);
	    // 尾部也要增加一个哨兵节点
	    heights.push_back(0);
	    for (int i = 0; i < heights.size(); i++)
	    {
	    	// 单调递增栈非空 and 或者遇到比栈顶元素小的元素
	        while (!st.empty() && heights[st.back()] > heights[i])
	        {
	            int cur = st.back(); // 要计算的柱体下标
	            st.pop_back();
	            int left = st.back() + 1;// 弹出当前下标后，现有栈的最大值的下标+1
	            int right = i - 1;// 当前柱体下标-1
	            // 宽=组成柱体的两边的下标相减+1
	            ans = max(ans, (right - left + 1) * heights[cur]);
	        }
	        st.push_back(i);
	    }
	    return ans;
	}
};
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36. 85. 最大矩形

// 广度优先遍历（DFS）
class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int m = matrix.size();
        if (m == 0) {
            return 0;
        }
        int n = matrix[0].size();
        // 使用二维数组left 记录，其中left[i][j] 为矩阵第 i 行第 j 列元素的左边连续 1 的数量
        vector<vector<int>> left(m, vector<int>(n, 0));

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '1') {
                    left[i][j] = (j == 0 ? 0: left[i][j - 1]) + 1;
                }
            }
        }

        int ret = 0;
        // 一直计算的都是以matrix[i][j] 为右下角的矩形
        for (int j = 0; j < n; j++) { // 对于每一列，使用基于柱状图的方法
            vector<int> up(m, 0), down(m, 0);

            stack<int> stk;
            for (int i = 0; i < m; i++) {
                while (!stk.empty() && left[stk.top()][j] >= left[i][j]) {
                    stk.pop();
                }
                up[i] = stk.empty() ? -1 : stk.top();
                stk.push(i);
            }
            stk = stack<int>();
            for (int i = m - 1; i >= 0; i--) {
                while (!stk.empty() && left[stk.top()][j] >= left[i][j]) {
                    stk.pop();
                }
                down[i] = stk.empty() ? m : stk.top();
                stk.push(i);
            }

            for (int i = 0; i < m; i++) {
                int height = down[i] - up[i] - 1;
                int area = height * left[i][j];
                ret = max(ret, area);
            }
        }
        return ret;
    }
};

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37. 94. 二叉树的中序遍历

// 二叉树的遍历
class Solution {
public:
    vector<int> result;

    void inorder(TreeNode* root){
        if(!root) return;

        inorder(root->left);
        result.push_back(root->val);
        inorder(root->right);
    }

    vector<int> inorderTraversal(TreeNode* root) {
        inorder(root);
        return result;
    }
};
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38. 96. 不同的二叉搜索树

// 动态规划
class Solution {
public:
	// dp[3],就是元素1的头节点搜索树的数量+元素2为头节点搜索树的数量+元素3为头节点搜索树的数量
	// dp[3] =  dp[2] * dp[0] + dp[1] * dp[1] + dp[0] * dp[2]
    int numTrees(int n) {
    	// dp[i]:1到i为节点组成的二叉搜索树的个数为dp[i]
        vector<int> dp(n+1);
        dp[0] = 1;
        for(int i{1};i<=n;++i){
            for(int j{1};j<=i;++j){
            	// dp[i] += dp[以j为头结点左子树节点数量] * dp[以j为头结点右子树节点数量],j相当于是头节点的元素，从1遍历到i为止
                dp[i] += dp[j-1]*dp[i-j];
            }
        }

        return dp[n];
    }
};
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39. 98. 验证二叉搜索树

// 二叉树的遍历
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> v;

    bool recursive(TreeNode* root){
        if(!root) return true;

        recursive(root->left);
        v.push_back(root->val);
        recursive(root->right);

        return true;
    }
    bool isValidBST(TreeNode* root) {
        recursive(root);

        for(int i{};i<v.size();++i){
            if(i && v[i-1] >= v[i]){
                return false;
            }
        }
        return true;
    }
};
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40. 101. 对称二叉树

// 二叉树递归
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool recursive(TreeNode* left,TreeNode* right){
        if(!left && !right) return true;
        else if(left && !right) return false;
        else if(!left && right) return false;

        if(left->val != right->val) return false;
        
        bool flag = recursive(left->left, right->right);
        bool flag1 = recursive(left->right, right->left);

        return flag && flag1;
    }
    bool isSymmetric(TreeNode* root) {
        if(!root) return true;

        return recursive(root->left,root->right);
    }
};
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41. 102. 二叉树的层序遍历

// 二叉树的遍历
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> result;

    vector<vector<int>> levelOrder(TreeNode* root) {
        if(!root) return {};

        queue<TreeNode*> q;
        q.push(root);

        while(!q.empty()){
            int n = q.size();
            vector<int> v1;
            for(int i{};i<n;++i){
                TreeNode* node = q.front();
                q.pop();

                v1.emplace_back(node->val);

                if(node->left) q.push(node->left);
                if(node->right) q.push(node->right);
            }

            result.push_back(v1);
        }
        return result;
    }
};
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42. 104. 二叉树的最大深度

// 树的递归
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        if(!root) return {};

        return max(maxDepth(root->left),maxDepth(root->right)) + 1;
    }
};
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43. 105. 从前序与中序遍历序列构造二叉树

// 树的遍历
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* recursive(vector<int>& inorder, vector<int>& preorder){
    	// 递归终止条件
        if(preorder.size() == 0) return {};
		// 后续遍历的最后一个是根节点
        TreeNode* root = new TreeNode(preorder[0]);
		
		// 在中序中找到根节点，左边的就是左子树、右边的数据就是右子树
        int index{};
        for(int i{};i<inorder.size();++i){
            if(inorder[i] == preorder[0]) index = i;
        }
        // 左闭右开的区间，然后拆分左右子树再次进行递归
        vector<int> leftInorder(inorder.begin(),inorder.begin() + index);
        vector<int> rightInorder(inorder.begin() + index +1,inorder.end());

        vector<int> leftPreorder(preorder.begin()+1,preorder.begin()+1+ leftInorder.size());
        vector<int> rightPreorder(preorder.begin()+1+ leftInorder.size(),preorder.begin()+1+ leftInorder.size() + rightInorder.size());

        root->left = recursive(leftInorder,leftPreorder);
        root->right = recursive(rightInorder,rightPreorder);

        return root;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.size() ==0 || inorder.size() == 0) return {};

        return recursive(inorder,preorder);
    }
};
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44. 114. 二叉树展开为链表

// 问题转化为寻找当前节点的前驱节点
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root) {
        if(!root) return;
        // 寻找当前节点的前驱节点
        TreeNode* cur = root;
        while(cur){
            while(cur->left){
                // 当前节点的左节点
                TreeNode* next = cur->left;
                TreeNode* pre  = next;
                // 得到前驱节点
                while(pre->right){
                    pre = next->right;
                }

                pre->right = cur->right;
                cur->left = nullptr;
                cur->right = next;
            }
            cur = cur->right;
        }
    }
};
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45. 121. 买卖股票的最佳时机

// 贪心
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int result{};
        // 记录最小值
        int minValue = prices[0];
        for(int i{1};i<prices.size();++i){
        	// 贪心获取一次买卖的最大利润
            result = max(result,(prices[i] - minValue));
            minValue = min(minValue,prices[i]);
        }
        return result;
    }
};
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46. 124. 二叉树中的最大路径和

// 递归求解（从根节点向上求解）
class Solution {
private:
    int maxSum = INT_MIN;

public:
    int maxGain(TreeNode* node) {
        if (node == nullptr) {
            return 0;
        }
        
        // 递归计算左右子节点的最大贡献值
        // 只有在最大贡献值大于 0 时，才会选取对应子节点
        int leftGain = max(maxGain(node->left), 0);
        int rightGain = max(maxGain(node->right), 0);

        // 节点的最大路径和取决于该节点的值与该节点的左右子节点的最大贡献值，因为路径不是自上而下的遍历
        int priceNewpath = node->val + leftGain + rightGain;

        // 更新答案
        maxSum = max(maxSum, priceNewpath);

        // 返回节点的最大贡献值
        return node->val + max(leftGain, rightGain);
    }

    int maxPathSum(TreeNode* root) {
        maxGain(root);
        return maxSum;
    }
};

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47. 128. 最长连续序列

// 哈希表+模拟
class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_set<int> num_set;
        for (const int& num : nums) {
            num_set.insert(num);
        }

        int longestStreak = 0;

        for (const int& num : num_set) {
        	// 核心关键是这个内层的跳过的逻辑
        	// 枚举的数 x 一定是在数组中不存在前驱数 x-1
            if (!num_set.count(num - 1)) {
                int currentNum = num;
                int currentStreak = 1;

                while (num_set.count(currentNum + 1)) {
                    currentNum += 1;
                    currentStreak += 1;
                }

                longestStreak = max(longestStreak, currentStreak);
            }
        }

        return longestStreak;           
    }
};

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48. 136. 只出现一次的数字

// 位运算
// 异或运算的性质
// 1.任何数和0做异或运算，结果仍是原来的数
// 2.任何数和自己异或结果是0
// 3.异或运算满足交换律和结合律
class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int n = nums.size();

        int result{nums[0]};

        for(int i{1};i<n;++i){
            result ^= nums[i];
        }

        return result;
    }
};
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49. 139. 单词拆分

// 动态规划、完全背包问题
class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> unSet(wordDict.begin(),wordDict.end());
        int n = s.size();
        // 1.确定dp数组以及下标的含义
        // dp[i]为true表示 字符串从1-i可以被拆分成多个单词
        vector<int> dp(n+1);
        dp[0] = true;

        for(int j = 1;j<=n;++j){ // 遍历背包
            for(int i{};i<=j;++i){ // 遍历物品
                string str = s.substr(i,j-i);

                if(dp[i] && unSet.find(str) != unSet.end()){
                    dp[j] = true;
                }
            }
        }
        return dp[n];
    }
};
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50. 141. 环形链表

// 双指针
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode *slow = head;
        ListNode *fast = head;

        while(fast && fast->next){
            slow = slow->next;
            fast = fast->next->next;

            if(slow == fast){
                return true;
            }
        }

        return false;
    }
};
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51. 142. 环形链表 II

// 双指针 + 数学
// 核心：a=c+(n−1)(b+c) 的等量关系，我们会发现：从相遇点到入环点的距离加上 n-1n−1 圈的环长，恰好等于从链表头部到入环点的距离
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(!head || !head->next || !head->next->next) return {};

        ListNode* slow = head;
        ListNode* fast = head;

        while(fast && fast->next){
            slow = slow->next;
            fast = fast->next->next;

            if(slow == fast){
                break;
            }
        }

        if(slow != fast) return{};

        slow = head;

        while(slow != fast){
            slow = slow->next;
            fast = fast->next;
        }

        return slow;
    }
};
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52. 146. LRU 缓存

// LRU(最近最少使用缓存)
// hashmap建立索引、list创建最近最少使用淘汰机制
class LRUCache {
public:
    LRUCache(int capacity) {
        size = capacity;
    }
    
    int get(int key) {
        if(m.find(key) == m.end()){
            return -1;
        }
        l.splice(l.begin(),l,m[key]);
        return m[key]->second;
    }
    
    void put(int key, int value) {
        if(m.find(key) != m.end()){
            l.splice(l.begin(),l,m[key]);
            m[key]->second = value;
            return;
        }

        if(size == l.size()){
            auto dKey = l.back().first;
            l.pop_back();
            m.erase(dKey);
        }

        l.push_front({key,value});
        m[key] = l.begin();
    }

private:
    list<pair<int,int>> l;
    unordered_map<int,list<pair<int,int>>::iterator> m;
    int size{};
};

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache* obj = new LRUCache(capacity);
 * int param_1 = obj->get(key);
 * obj->put(key,value);
 */
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53. 148. 排序链表

// 链表

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        vector<int> v;

        ListNode* cur = head;

        while(cur){
            v.push_back(cur->val);
            cur = cur->next;
        }

        sort(v.begin(),v.end());
        cur = head;
        for(int i{};i<v.size();++i){
            cur->val = v[i];
            cur = cur->next;
        }

        return head;
    }
};
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54. 152. 乘积最大子数组

// 动态规划
class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int ans{INT_MIN},n = nums.size();

        if(1==n) return nums[0];

        vector<int> dpMax(n);
        vector<int> dpMin(n);

        dpMax[0] = nums[0];
        dpMin[0] = nums[0];
        ans = nums[0];
        for(int i = 1;i<n;++i){
            dpMax[i] = max(nums[i],max(nums[i]*dpMax[i-1],nums[i]*dpMin[i-1]));
            dpMin[i] = min(nums[i],min(nums[i]*dpMax[i-1],nums[i]*dpMin[i-1]));
            ans = max(ans,dpMax[i]);
        }

        return ans;
    }
};
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55. 155. 最小栈

// 栈
// 使用两个栈，专门维护一个储存最小值的栈
class MinStack {
public:
    MinStack() {

    }
    
    void push(int val) {
        if(s2.empty() || val <= s2.top()) s2.push(val);
        s1.push(val);
    }
    
    void pop() {
        if(s1.empty() && s2.empty()) return;
        if(s1.top() == s2.top()) s2.pop();
        s1.pop();
    }
    
    int top() {
        if(s1.empty()) return -1;
        return s1.top();
    }
    
    int getMin() {
        if(s2.empty()) return -1;
        return s2.top();
    }
private:
    stack<int> s1;
    stack<int> s2;
};

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack* obj = new MinStack();
 * obj->push(val);
 * obj->pop();
 * int param_3 = obj->top();
 * int param_4 = obj->getMin();
 */
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56. 160. 相交链表

// 链表+哈希表
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        unordered_set<ListNode*> unS;

        while(headA){
            unS.insert(headA);
            headA = headA->next;
        }

        while(headB){
            if(unS.find(headB)!=unS.end()) return headB;
            headB = headB->next;
        }

        return {};
    }
};
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57. 169. 多数元素

// 哈希表
class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int,int> mp;

        for(int num : nums) {
            mp[num]++;
            if(mp[num] > nums.size()/2) return num;
            
        }

        return -1;
    }
};
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58. 198. 打家劫舍

// 动态规划
class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        // 1.dp数组以及下标定义
        // dp[i][0] 第i-1天不偷窃，最多可以偷窃的金钱的数目为dp[i][0]
        // dp[i][1] 第i-1天偷窃
        vector<vector<int>> dp(n,vector<int>(2));

        dp[0][0] = 0;
        dp[0][1] = nums[0];

        int result{};

        for(int i{1};i<n;++i){
            dp[i][0] = max(dp[i-1][0],dp[i-1][1]);
            dp[i][1] = max(dp[i-1][0],dp[i-1][0] + nums[i]);
        }

        return max(dp[n-1][0],dp[n-1][1]);
    }
};
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59. 200. 岛屿数量

// dfs广度有限遍历
class Solution {
public:
    void dfs(int i,int j,vector<vector<char>>& grid){
        if(min(i,j) < 0 || i>=grid.size() || j >= grid[0].size()) return;
		
		// 如果是水或者已经被访问过，即是其它的小岛的组成部分就跳过
        if(grid[i][j] == '0' || grid[i][j] == '2') return;
        grid[i][j] = '2';
		// 广度优先遍历
        for(int d{};d<4;++d){
            dfs(i+dx[d],j+dy[d],grid);
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int count{};
		// 依次进行广搜
        for(int i{};i<m;++i){
            for(int j{};j<n;++j){
                if(grid[i][j] == '1'){
                    dfs(i,j,grid);
                    count++;
                }
            }
        }

        return count;
    }

private:
    int dx[4]{1,0,-1,0};
    int dy[4]{0,-1,0,1};
};
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60. 206. 反转链表

// 双指针 + 画图

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(!head) return {};

        ListNode* pre = nullptr;
        ListNode* cur = head;

        while(cur){
            ListNode* node = cur->next;
            cur->next = pre;
            pre = cur;
            cur = node;
        }

        return pre;
    }
};
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61. 207. 课程表

// 拓扑排序+dfs+bfs
class Solution {
private:
	// 表示有向有环图
    vector<vector<int>> edges;
    // 记录每个节点的入度
    vector<int> indeg;

public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        edges.resize(numCourses);
        indeg.resize(numCourses);
        for (const auto& info: prerequisites) {
            edges[info[1]].push_back(info[0]);
            ++indeg[info[0]];
        }

        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.push(i);
            }
        }

        int visited = 0;
        while (!q.empty()) {
            ++visited;
            int u = q.front();
            q.pop();
            for (int v: edges[u]) {
                --indeg[v];
                if (indeg[v] == 0) {
                    q.push(v);
                }
            }
        }

        return visited == numCourses;
    }
};
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62. 208. 实现 Trie (前缀树)

// 字典树
class Trie {
private:
    vector<Trie*> children;
    bool isEnd; // 表示该字符是否字符串的结尾

    Trie* searchPrefix(string prefix) {
        Trie* node = this;
        for (char ch : prefix) {
            ch -= 'a';
            if (node->children[ch] == nullptr) {
                return nullptr;
            }
            node = node->children[ch];
        }
        return node;
    }

public:
    Trie() : children(26), isEnd(false) {}

    void insert(string word) {
        Trie* node = this;
        for (char ch : word) {
            ch -= 'a';
            if (node->children[ch] == nullptr) {
                node->children[ch] = new Trie();
            }
            node = node->children[ch];
        }
        node->isEnd = true;
    }

    bool search(string word) {
        Trie* node = this->searchPrefix(word);
        return node != nullptr && node->isEnd;
    }

    bool startsWith(string prefix) {
        return this->searchPrefix(prefix) != nullptr;
    }
};
wor
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63. 215. 数组中的第K个最大元素

// 排序+优先队列
class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        //使用其它算法中的sort算法进行排序
        // sort(nums.begin(),nums.end(),[&](auto &&num1,auto &&num2){
        //     return num1 > num2;
        // });

        // return nums[k-1];
        //使用优先队列，大根堆
        priority_queue<int> pq(nums.begin(),nums.end());

        for(int i{};i<k;++i){
            if(i==k-1) return pq.top();
            pq.pop();
        }
        return -1;
    }
};
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64. 221. 最大正方形

// 动态规划
class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        if(matrix.size() == 0 || matrix[0].size() == 0)return 0;
        int m = matrix.size();
        int n = matrix[0].size();
        int maxNum{};
        // dp(i,j) 表示以 (i,j) 为右下角，且只包含 1 的正方形的边长最大值
        vector<vector<int>> dp(m,vector<int>(n,0));

        for(int i{};i<m;++i){
            for(int j{};j<n;++j){
                if(matrix[i][j] == '1'){
                    if(min(i,j) == 0) dp[i][j] = 1;
                    else dp[i][j] = min(min(dp[i-1][j-1],dp[i][j-1]),dp[i-1][j]) +1;
                }
                maxNum = max(maxNum,dp[i][j]);
            }
        }
        return maxNum*maxNum;
    }
};
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65. 226. 翻转二叉树

// 递归
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invert(TreeNode* root){
        if(!root) return root;

        
        TreeNode* left = root->left;
        TreeNode* right = root->right;

        root->right = left;
        root->left = right;
        if(root->right) invert(root->right);
        if(root->left) invert(root->left);

        return root;
    }
    TreeNode* invertTree(TreeNode* root) {
        return invert(root);
    }
};
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66. 234. 回文链表

// 双指针 + 链表
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        vector<int> v;

        while(head){
            v.push_back(head->val);
            head = head->next;
        }

        int left{};
        int right = v.size()-1;

        while(left < right){
            if(v[left] != v[right]){
                return false;
            }
            left++;
            right--;
        }

        return true;
    }
};
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67. 236. 二叉树的最近公共祖先

// 递归
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* recursive(TreeNode* root,TreeNode* p,TreeNode* q){
        if(root == p || root == q|| !root) return root;

        TreeNode* left  = recursive(root->left,p,q);
        TreeNode* right = recursive(root->right,p,q);

        if(left && right) return root;
        else if( left && !right) return left;
        else if( !left && right) return right;
        else return {}; 
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        return recursive(root,p,q);
    }
};

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68. 238. 除自身以外数组的乘积

// 前缀和
class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        //前缀和。i左侧与右侧的乘积相乘
        int n = nums.size();
        vector<int> l(n),r(n),ans(n);
        l[0] = nums[0],r[n-1] = nums[n-1];

        for(int i = 1;i<n;++i) l[i] = l[i-1]*nums[i];  
        for(int i = n-2;i>=0;--i) r[i] = r[i+1]*nums[i];

        ans[0] = r[1];
        ans[n-1] = l[n-2];
        for(int i = 1;i<n-1;++i){
            ans[i] = l[i-1]*r[i+1];
        }

        return ans;
    }
};
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69. 239. 滑动窗口最大值

// 滑动窗口
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        int n = nums.size();
        priority_queue<pair<int,int>> pq;
        for(int i{};i<k;++i) pq.push(pair(nums[i],i));

        vector<int> result;
        result.push_back(pq.top().first);

        for(int i{k};i<n;++i){
            pq.push(pair(nums[i],i));

            while(pq.top().second <= i-k){
                pq.pop();
            }

            result.push_back(pq.top().first);
        }

        return result;
    }
};
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70. 240. 搜索二维矩阵 II

// 二分查找
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(),n = matrix[0].size();

        for(int i{};i<m;++i){
        	// 先找数组中第一个不小于target的元素（upper_bound是找数组中大于目标值的第一个元素）
            auto iter = lower_bound(matrix[i].begin(),matrix[i].end(),target);

            if(iter != matrix[i].end() && *iter == target) return true;
        }
        return false;
    }
};
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71. 253. 会议室 II

// 优先队列（最小堆）
class Solution {
public:
    int minMeetingRooms(vector<vector<int>>& intervals) {
        sort(intervals.begin(),intervals.end());

        priority_queue<int,vector<int>,greater<int>> pq;
        for(auto &meet : intervals){
            if(!pq.empty() && pq.top() <= meet[0]){
                pq.pop();
            }
            pq.push(meet[1]);
        }

        return pq.size();
    }
};
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72. 279. 完全平方数

// 动态规划
class Solution {
public:
    int numSquares(int n) {
    	// 1.dp数组以及数组下标定义
    	// dp[i]表示和为i的完全平方数的最少数量
        vector<int> dp(n+1,INT_MAX);    
        dp[0] = 0;

        for(int i = 1;i*i<= n;++i){
            for(int j = 1;j<=n;++j){
            	// 2.确定递推公式
                if(j-i*i >= 0) dp[j] = min(dp[j],dp[j-i*i] + 1);
            }
        }

        return dp[n] == INT_MAX ? -1 : dp[n];
    }
};
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73. 283. 移动零

// 双指针
class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int left{},right{};
        for(;right < nums.size();++right){
            if(nums[right]) swap(nums[right],nums[left++]);
        }
    }
};
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74. 287. 寻找重复数

// 二分查找
class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int left{};
        int n = nums.size();
        int right = n - 1;
        int ans{};
        while(left <= right){
            int mid = (left + right)/2;
            // 计算数组中<=mid的个数
            int count{};
            for(int i{};i<n;++i) count += nums[i] <= mid;

            if(count > mid){
                right = mid-1;
                ans = mid;
            }else{
                left = mid+1;
            }

            
        }
        return ans;
    }
};
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75. 297. 二叉树的序列化与反序列化

// 二叉树的遍历
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Codec {
public:
	// 前序遍历编码
    void preorder(TreeNode* root,string &s){
        if(!root) {
            s+=" null";
            return;
        }

        s+= (" " + to_string(root->val));
        preorder(root->left,s);
        preorder(root->right,s);
    }
    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        string s;
        preorder(root,s);
        return s;
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        istringstream is(data);

        return buildNode(is);
    }
	// 根据前序遍历解码
    TreeNode* buildNode(istringstream& is){
        string s;
        is >> s;

        if (s == "null")
			return NULL;

        TreeNode* node = new TreeNode(stoi(s));
        node->left = buildNode(is);
        node->right = buildNode(is);

        return node;
    }
};

// Your Codec object will be instantiated and called as such:
// Codec ser, deser;
// TreeNode* ans = deser.deserialize(ser.serialize(root));
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76. 300. 最长递增子序列

// 动态规划
class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        if(n <= 1) return n;
        vector<int> dp(n,1);
        
        int result{};
        for(int i = 1;i<n;++i){
            for(int j{};j<i;++j){
                if(nums[i] > nums[j]) dp[i] = max(dp[i],dp[j] + 1);
            }

            result = max(dp[i],result);
        }

        return result;
    }
};
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77. 301. 删除无效的括号

// 深度优先遍历（DFS）
class Solution {
public:
    vector<string> res;
    // 判断一个括号是否合法
    inline bool isValid(string &s){
        int result{};

        for(int i{};i<s.size();++i){
            if(s[i] == '('){
                result++;
            }else if(s[i] == ')'){
                result--;

                if(result < 0) return false;
            }
        }

        return result == 0;
    }

    void helper(string str,int start,int lremove,int rremove){
        if (lremove == 0 && rremove == 0) {
            if (isValid(str)) {
                res.push_back(str);
            }
            return;
        }
        for (int i = start; i < str.size(); i++) {
            if (i != start && str[i] == str[i - 1]) {
                continue;
            }
            // 如果剩余的字符无法满足去掉的数量要求，直接返回
            if (lremove + rremove > str.size() - i) {
                return;
            } 
            // 尝试去掉一个左括号
            if (lremove > 0 && str[i] == '(') {
                helper(str.substr(0, i) + str.substr(i + 1), i, lremove - 1, rremove);
            }
            // 尝试去掉一个右括号
            if (rremove > 0 && str[i] == ')') {
                helper(str.substr(0, i) + str.substr(i + 1), i, lremove, rremove - 1);
            }
        }

    }
    vector<string> removeInvalidParentheses(string s) {
        int lremove{},rremove{};

        for(char &c : s){
            if(c == '('){
                lremove++;
            }else if(c == ')'){
                if(lremove == 0){
                    rremove++;
                }else{
                    lremove--;
                }
            }
        }

        helper(s,0,lremove,rremove);

        return res;
    }
};
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78. 309. 最佳买卖股票时机含冷冻期

// 动态规划
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        // 0：状态一，买入股票
        // 1：状态二，卖出了股票，度过了冷冻期
        // 2：状态三，今天卖出了股票，还在冷冻期
        // 3：状态四，今天为冷冻期状态，但冷冻期状态不可持续，只有一天

        vector<vector<int>> dp(n,vector<int>(4));

        dp[0][0] = -prices[0];

        for(int i = 1; i < n;++i){
            dp[i][0] = max(dp[i-1][0],max(dp[i-1][3],dp[i-1][1]) - prices[i]);
            dp[i][1] = max(dp[i-1][1],dp[i-1][3]);
            dp[i][2] = dp[i-1][0] + prices[i];
            dp[i][3] = dp[i-1][2];
        }

        return max(dp[n-1][1],max(dp[n-1][2],dp[n-1][3]));
    }
};
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79. 312. 戳气球

// 动态规划
class Solution {
public:
    int maxCoins(int[] nums) {
	    int n = nums.length;
	    // 添加两侧的虚拟气球
	    int[] points = new int[n + 2];
	    points[0] = points[n + 1] = 1;
	    for (int i = 1; i <= n; i++) {
	        points[i] = nums[i - 1];
	    }
	    // base case 已经都被初始化为 0
	    // dp[i][j] = x表示，戳破气球i和气球j之间（开区间，不包括i和j）的所有气球，可以获得的最高分数为x
	    int[][] dp = new int[n + 2][n + 2];
	    // 开始状态转移
	    // i 应该从下往上
	    for (int i = n; i >= 0; i--) {
	        // j 应该从左往右
	        for (int j = i + 1; j < n + 2; j++) {
	            // 最后戳破的气球是哪个？
	            for (int k = i + 1; k < j; k++) {
	                // 择优做选择
	                dp[i][j] = Math.max(
	                    dp[i][j], 
	                    dp[i][k] + dp[k][j] + points[i]*points[j]*points[k]
	                );
	            }
	        }
	    }
	    return dp[0][n + 1];
	}
};
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80. 322. 零钱兑换

// 动态规划
class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> dp(amount+1,INT_MAX);

        dp[0] = 0;
		// 外层遍历物品
        for(int i{};i<coins.size();++i){
        	// 内层遍历背包容量
            for(int j = coins[i];j<=amount;++j){
                if(dp[j-coins[i]] < INT_MAX)dp[j] = min(dp[j],dp[j-coins[i]] + 1);
            }
        }

        return dp[amount] == INT_MAX ? -1 : dp[amount];
    }
};
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81. 337. 打家劫舍 III

// 动态规划+递归
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> recursive(TreeNode* root){
        if(!root) return vector<int>{0, 0};

        vector<int> left = recursive(root->left);
        vector<int> right = recursive(root->right);

        // 偷
        int v1 = root->val + left[0] + right[0];
        // 不偷
        // 如果不偷当前节点，那么左右孩子就可以偷，至于到底偷不偷一定是选一个最大的，所以：val2 = max(left[0], left[1]) + max(right[0], right[1]);
        int v2 = max(left[0],left[1]) + max(right[0],right[1]);

        return {v2,v1};
    }
    int rob(TreeNode* root) {
    	// dp数组以及下标的含义：下标为0记录不偷该节点所得到的的最大金钱，下标为1记录偷该节点所得到的的最大金钱
        vector<int> result = recursive(root);
        return max(result[0],result[1]);
    }
};
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82. 338. 比特位计数

// 位运算
class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> result;

        for(int i{};i<=n;++i){
            int count{},num = i;
            // 求位1的个数
            while(num > 0){
                if(num&1) ++count;
                num = num >> 1;
            }
            result.push_back(count);
        }
        return result;
    }
};
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83. 347. 前 K 个高频元素

// 优先队列
class mycomparison{
    public:
    bool operator()(const pair<int,int>&lhs,const pair<int,int> &rhs){
        return lhs.second > rhs.second;
    }
};
class Solution {
public:
    
    vector<int> topKFrequent(vector<int>& nums, int k) {
        // 获取频率
        unordered_map<int,int> unMap;
        for(int &num : nums) unMap[num]++;
        // 构建小顶堆
        priority_queue<pair<int,int>,vector<pair<int,int>>,mycomparison> pq;

        for(auto &[key,value] : unMap){
            pq.push(pair(key,value));

            if(pq.size() > k){
                pq.pop();
            }
        }

        // 返回答案
        vector<int> result;
        while(!pq.empty()){
            result.emplace_back(pq.top().first);
            pq.pop();
        }

        return result;
    }
};
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84. 394. 字符串解码

// 栈（模拟栈）
class Solution {
public:
    string getDigits(string &s, size_t &ptr) {
        string ret = "";
        while (isdigit(s[ptr])) {
            ret.push_back(s[ptr++]);
        }
        return ret;
    }

    string getString(vector <string> &v) {
        string ret;
        for (const auto &s: v) {
            ret += s;
        }
        return ret;
    }

    string decodeString(string s) {
        vector <string> stk;
        size_t ptr = 0;

        while (ptr < s.size()) {
            char cur = s[ptr];
            if (isdigit(cur)) {
                // 获取一个数字并进栈
                string digits = getDigits(s, ptr);
                stk.push_back(digits);
            } else if (isalpha(cur) || cur == '[') {
                // 获取一个字母并进栈
                stk.push_back(string(1, s[ptr++])); 
            } else {
                ++ptr;
                vector <string> sub;
                while (stk.back() != "[") {
                    sub.push_back(stk.back());
                    stk.pop_back();
                }
                reverse(sub.begin(), sub.end());
                // 左括号出栈
                stk.pop_back();
                // 此时栈顶为当前 sub 对应的字符串应该出现的次数
                int repTime = stoi(stk.back()); 
                stk.pop_back();
                string t, o = getString(sub);
                // 构造字符串
                while (repTime--) t += o; 
                // 将构造好的字符串入栈
                stk.push_back(t);
            }
        }

        return getString(stk);
    }
};
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85. 399. 除法求值

// 并查集
class UnionFind {
private:
    vector<int> parent; // 存放父节点
    vector<double> weight; // 指向父节点的权值
    
public:
    UnionFind(int n) {
        for (int i = 0; i < n; ++i) {
            parent.push_back(i);
            weight.push_back(1.0); // 权重初始化为1
        }
    }
    // 路径压缩。返回根节点id
    int find(int x) {
        // 递归寻找根节点，更新该点到根的权重为该点父节点到根的权重
        if (x != parent[x]) {
            int origin = parent[x];
            parent[x] = find(parent[x]);
            weight[x] *= weight[origin];
        }
        return parent[x];
    }
    // 返回除法结果。如果两个值不存在则-1
    double isConected(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        // 如果两个值有共同的根也就是可以计算，则算结果。否则不在同一个并查集，-1
        if (rootX == rootY) {
            return weight[x] / weight[y];
        } else {
            return -1.00000;
        }
    }
    void myunion(int x, int y, double value) {
        // 分别找到二者的根节点
        int rootX = find(x), rootY = find(y);
        if (rootX == rootY) {
            return; // 二者已经指向同一个根节点
        }
        // 令分子指向分母的根节点，权重为分母到根的权重*分母除分子的值/分子到根的权重。一开始都是1
        parent[rootX] = rootY;
        weight[rootX] = weight[y] * value / weight[x];
    }
};
class Solution {
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        // 初始化并查集
        int equationsSize = equations.size();
        UnionFind unionFind(2 * equationsSize);
        // 第 1 步：预处理，将变量的值与 id 进行映射
        map<string, int> hashMap;
        int id = 0;
        for (int i = 0; i < equationsSize; ++i) {
            // 存分子，分母，值为id
            vector<string> equation = equations[i];
            string var1 = equation[0];
            string var2 = equation[1];
            if (!hashMap.count(var1)) {
                hashMap[var1] = id;
                ++id;
            }
            if (!hashMap.count(var2)) {
                hashMap[var2] = id;
                ++id;
            }
            // 把分子分母用有向边连起来
            unionFind.myunion(hashMap[var1], hashMap[var2], values[i]);
        }
        // 第 2 步：做查询
        int queriesSize = queries.size();
        vector<double> res(queriesSize, -1.00000);
        for (int i = 0; i < queriesSize; ++i) {
            string var1 = queries[i][0];
            string var2 = queries[i][1];
            int id1, id2;
            // 如果两个值有至少一个不在equations中，结果为-1，否则做除法
            if (hashMap.count(var1) && hashMap.count(var2)) {
                id1 = hashMap[var1];
                id2 = hashMap[var2];
                res[i] = unionFind.isConected(id1, id2);
            }
        }
        return res;
    }
};
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86. 406. 根据身高重建队列

// 贪心
class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(),people.end(),[&](auto &&a,auto &&b){
            if(a[0] == b[0]) return a[1] < b[1];
            return a[0] > b[0];
        });

        vector<vector<int>> result;
        for(int i{};i<people.size();++i){
            result.insert(result.begin() + people[i][1],people[i]);
        }

        return result;
    }
};
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87. 416. 分割等和子集

// 动态规划 01背包问题
// 一个商品如果可以重复多次放入是完全背包，而只能放入一次是01背包
class Solution {
public:
    bool canPartition(vector<int>& nums) {
    	// dp[j]表示 背包总容量是j，最大可以凑成j的子集总和为dp[j]
        int nSum = accumulate(nums.begin(),nums.end(),0);

        if(nSum %2 != 0 ) return false;

        int target = nSum/2;
        int n = nums.size();
        vector<int> dp(target+1);
		// 如果使用一维dp数组，物品遍历的for循环放在外层，遍历背包的for循环放在内层，且内层for循环倒序遍历
        for(int i{};i<n;++i){
            for(int j = target;j >=nums[i];--j){
                dp[j] = max(dp[j],dp[j-nums[i]] + nums[i]);
            }
        }

        return dp[target] == target;
    }
};
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88. 437. 路径总和 III

// 暴力递归
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    long long rootSum(TreeNode* root, long long targetSum){
        if(!root) return{};
        long long ret{};
        if(targetSum == root->val) ret++;
        ret+= rootSum(root->left,targetSum-root->val);
        ret+= rootSum(root->right,targetSum-root->val);
        return ret;
    }
    int pathSum(TreeNode* root, int targetSum) {
        if(!root) return{};
        long long ret{};

        ret = rootSum(root,targetSum);

        ret+= pathSum(root->left,targetSum);
        ret+= pathSum(root->right,targetSum);

        return (int)ret;
    }
};
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89. 438. 找到字符串中所有字母异位词

// 滑动窗口
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        int slen = s.size();
        int plen = p.size();

        if(slen < plen) return{};

        vector<int> sArr(26);
        vector<int> pArr(26);
        vector<int> ans;

        for(char c : p){
            pArr[c - 'a']++;
        }

        for(int i{};i<plen;++i){
            sArr[s[i] - 'a']++;
        }

        if(sArr == pArr) ans.emplace_back(0);

        for(int i = 0;i< slen - plen;++i){
            sArr[s[i] - 'a']--;
            sArr[s[i + plen] - 'a']++;

            if(sArr == pArr) ans.emplace_back(i+1);
        }

        return ans;
    }
};
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90. 448. 找到所有数组中消失的数字

// 哈希表
class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        unordered_map<int,int> unMap;

        vector<int> result;

        for(int i{};i<nums.size();++i){
            unMap[nums[i]]++;
        }

        for(int i{1};i<=nums.size();++i){
            if(unMap[i] == 0) result.push_back(i);
        }

        return result;
    }
};
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91. 461. 汉明距离

//1.内置位计数功能
class Solution {
public:
    int hammingDistance(int x, int y) {
        return __builtin_popcount(x ^ y);
    }
};
// 位运算
//2.移位实现位计数
class Solution {
public:
    int hammingDistance(int x, int y) {
        int s = x ^ y, ret = 0;
        while (s) {
            ret += s & 1;
            s >>= 1;
        }
        return ret;
    }
};

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92. 494. 目标和

// 动态规划+01背包
class Solution {
public:
	// left - right = target
	// left + right = sum
	// left = (target + sum)/2
    int findTargetSumWays(vector<int>& nums, int target) {
        int nSum = accumulate(nums.begin(),nums.end(),0);

        if(nSum < abs(target)) return 0;

        int target1 = (nSum + target)/2;
        if((nSum + target)%2) return 0;
		//填满j（包括j）这么大容积的包，有dp[j]种方法
        vector<int> dp(target1+1);

        dp[0] = 1;

        for(int i{};i<nums.size();++i){
            for(int j = target1;j>=nums[i];--j){
                dp[j] += dp[j-nums[i]];
            }
        }

        return dp[target1];
    }
};
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93. 538. 把二叉搜索树转换为累加树

// 二叉树的遍历（后续遍历的变形）
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int pre{};
    void recursive(TreeNode* root){
        if(!root) return;

        recursive(root->right);
        root->val+=pre;
        pre = root->val;
        recursive(root->left);
    }
    TreeNode* convertBST(TreeNode* root) {
        recursive(root);

        return root;
    }
};
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94. 543. 二叉树的直径

// dfs深度优先搜索
class Solution {
    int ans;
    int depth(TreeNode* rt){
        if (rt == NULL) {
            return 0; // 访问到空节点了，返回0
        }
        int L = depth(rt->left); // 左儿子为根的子树的深度
        int R = depth(rt->right); // 右儿子为根的子树的深度
        ans = max(ans, L + R + 1); // 计算d_node即L+R+1 并更新ans
        return max(L, R) + 1; // 返回该节点为根的子树的深度
    }
public:
    int diameterOfBinaryTree(TreeNode* root) {
        ans = 1;
        depth(root);
        return ans - 1;
    }
};

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95. 560. 和为 K 的子数组

// 前缀和
class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int n = nums.size();
        unordered_map<int,int> unMap;
        unMap[0] = 1;

        vector<int> pre(n+1);
        int result{};
        for(int i{};i<n;++i){
            pre[i+1] = pre[i] + nums[i];
            if(unMap.find(pre[i+1]-k) != unMap.end()){
                result += unMap[pre[i+1]-k];
            }
            unMap[pre[i+1]]++;
        }

        return result;
    }
};
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96. 581. 最短无序连续子数组

// 排序
class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
        if (is_sorted(nums.begin(), nums.end())) {
            return 0;
        }
        vector<int> numsSorted(nums);
        sort(numsSorted.begin(), numsSorted.end());
        int left = 0;
        while (nums[left] == numsSorted[left]) {
            left++;
        }
        int right = nums.size() - 1;
        while (nums[right] == numsSorted[right]) {
            right--;
        }
        return right - left + 1;
    }
};

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97. 617. 合并二叉树

// 递归 深度优先搜索
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* recursive(TreeNode* root1,TreeNode* root2){
        if(!root1 && !root2) return {};
        else if(!root1 && root2) return root2;
        else if(root1 && !root2) return root1;

        root1->val += root2->val;
        root1->left = recursive(root1->left,root2->left);
        root1->right = recursive(root1->right,root2->right);
        return root1;
    }
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        return recursive(root1,root2);
    }
};
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98. 621. 任务调度器

// 桶思想
class Solution {
public:
    // 总排队时间 = (桶个数 - 1) * (n + 1) + 最后一桶的任务数
    int leastInterval(vector<char>& tasks, int n) {
        // 总的任务数
        int len=tasks.size();
        vector<int> vec(26);
        for(char c:tasks) ++vec[c-'A'];
        sort(vec.rbegin(),vec.rend());
        int cnt=1;
        // 得出最后一桶的任务数
        while(cnt<vec.size()&&vec[cnt]==vec[0]) cnt++;
        // vec[0] - 1 桶的数量
        // n+1 一个等待时间
        // cnt 最后一桶的任务数
        return max(len,cnt+(n+1)*(vec[0]-1) );
    }
};
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99. 647. 回文子串

// 动态规划
class Solution {
public:
    int countSubstrings(string s) {
        int len = s.size();
        int result{};
        vector<vector<bool>> dp(len,vector<bool>(len));
		// 从下到上、从左至右的遍历
        for(int i = len-1;i>=0;--i){
            for(int j = i;j<len;++j){
                if(s[i] == s[j]){
                    if(j-i <= 1){
                        result++;
                        dp[i][j] = true;
                    }else if(dp[i+1][j-1]){
                        result++;
                        dp[i][j] = true;
                    }
                }
            }
        }

        return result;
    }
};
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100. 739. 每日温度

// 单调栈
class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int n = temperatures.size();
        vector<int> ans(n);
        stack<int> s;
        for (int i = 0; i < n; ++i) {
            while (!s.empty() && temperatures[i] > temperatures[s.top()]) {
                int previousIndex = s.top();
                ans[previousIndex] = i - previousIndex;
                s.pop();
            }
            s.push(i);
        }
        return ans;
    }
};

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