1. 摆动排序II

给你一个整数数组 nums，将它重新排列成 nums[0] < nums[1] > nums[2] < nums[3]... 的顺序。

你可以假设所有输入数组都可以得到满足题目要求的结果。

Example

输入：nums = [1,5,1,1,6,4]
输出：[1,6,1,5,1,4]
解释：[1,4,1,5,1,6] 同样是符合题目要求的结果，可以被判题程序接受。
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代码

class Solution {
public:
    void wiggleSort(vector<int> &nums) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        vector<int> target(nums.size());
        int x = (n + 1) >> 1;
        for (int i = 0, j = x - 1, k = n - 1; i < n; i += 2, j--, k--) {
            target[i] = nums[j];
            if (i + 1 < n) target[i + 1] = nums[k];
        }
        nums = move(target);
    }
};
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2. 个位数字为K的整数之和

Given two integers num and k, consider a set of positive integers with the following properties:

• The units digit of each integer is k.
• The sum of the integers is num.

Return the minimum possible size of such a set, or -1 if no such set exists.

Note:

• The set can contain multiple instances of the same integer, and the sum of an empty set is considered 0.
• The units digit of a number is the rightmost digit of the number.

Example

输入：num = 58, k = 9
输出：2
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代码 [贪心]

class Solution {
public:
    int minimumNumbers(int num, int k) {
        if (num == 0) return 0;
        if ((num & 1) && !(k & 1)) return -1;
        int cnt[10]{0}, acc = 0, ans = 0, target = num % 10; // cnt 可用位运算
        while (acc <= num) {
            acc += k;
            if (cnt[acc % 10] == 1) return -1;
            ++ans;
            cnt[acc % 10] = 1;
            if (acc % 10 == target) return ans;
        }
        return -1;
    }
};
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3. 矩阵的最大非负积

You are given a m x n matrix grid. Initially, you are located at the top-left corner (0, 0), and in each step, you can only move right or down in the matrix.

Among all possible paths starting from the top-left corner (0, 0) and ending in the bottom-right corner (m - 1, n - 1), find the path with the maximum non-negative product. The product of a path is the product of all integers in the grid cells visited along the path.

Return the maximum non-negative product modulo 10^9 + 7. If the maximum product is negative, return -1.

Notice that the modulo is performed after getting the maximum product.

Example

输入：n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4
输出：0.16666666666666666 
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代码 [动态规划]

class Solution {
private:
    long long dot(long long x, long long y) {
        if (y == LONG_LONG_MIN) {
            return x > 0 ? LONG_LONG_MIN : LONG_LONG_MAX;
        }
        if (y == LONG_LONG_MAX) {
            return x > 0 ? LONG_LONG_MAX : LONG_LONG_MIN;
        }
        return x * y;
    }

public:
    int maxProductPath(const vector<vector<int>> &grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector<pair<long long, long long>>> dp(n, vector<pair<long long, long long>>(m, {LLONG_MAX, LLONG_MIN}));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (i == 0 && j == 0) {
                    dp[0][0] = {grid[0][0], grid[0][0]};
                    continue;
                }
                if (i > 0) {
                    long long t1 = dot(grid[i][j], dp[i - 1][j].first), t2 = dot(grid[i][j], dp[i - 1][j].second);
                    dp[i][j].first = min({dp[i][j].first, t1, t2});
                    dp[i][j].second = max({dp[i][j].second, t1, t2});
                }
                if (j > 0) {
                    long long t1 = dot(grid[i][j], dp[i][j - 1].first), t2 = dot(grid[i][j], dp[i][j - 1].second);
                    dp[i][j].first = min({dp[i][j].first, t1, t2});
                    dp[i][j].second = max({dp[i][j].second, t1, t2});
                }
            }
        }
        return dp[n - 1][m - 1].second < 0 ? -1 : dp[n - 1][m - 1].second % 1000000007ll;
    }
};
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4. 课程表

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

• For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return true if you can finish all courses. Otherwise, return false.

Example

输入：numCourses = 2, prerequisites = [[1,0]]
输出：true
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代码 [DFS]

class Solution {
private:
    enum {LEARNING, LEARNED, UNLEARNED};
    vector<int> status;
    vector<vector<int>> graph;

    bool dfs(int i) {
        status[i] = LEARNING;
        for (int next:graph[i]) {
            if (status[next] == LEARNED) continue;
            if (status[next] == LEARNING) return false;
            if (!dfs(next)) return false;
        }
        status[i] = LEARNED;
        return true;
    }

public:
    bool canFinish(int n, const vector<vector<int>> &prerequisites) {
        status.assign(n, UNLEARNED);
        graph.assign(n, {});
        for (auto &p:prerequisites) graph[p[0]].push_back(p[1]);
        for (int i = 0; i < n; ++i) {
            if (status[i] == LEARNED) continue;
            if (!dfs(i)) return false;
        }
        return true;
    }
};
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4. 路径总和III

给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之和等于 targetSum 的 路径 的数目。

路径 不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。

Example

输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出：3
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代码 [暴力DFS]

class Solution {
private:
    int result = 0, target;
    using ll = long long;

    unordered_map<ll, ll> map_union(unordered_map<ll, ll> m1, unordered_map<ll, ll> m2, int add) {
        unordered_map<ll, ll> ans;
        for (auto[x, y]:m1) ans[x + add] += y;
        for (auto[x, y]:m2) ans[x + add] += y;
        ans[add]++;
        return ans;
    }

    unordered_map<ll, ll> dfs(TreeNode *root) {
        unordered_map<ll, ll> left = root->left ? dfs(root->left) : unordered_map<ll, ll>();
        unordered_map<ll, ll> right = root->right ? dfs(root->right) : unordered_map<ll, ll>();
        unordered_map<ll, ll> ans = move(map_union(left, right, root->val));
        result += ans[target];
        return ans;
    }

public:
    int pathSum(TreeNode *root, int targetSum) {
        if (!root) return 0;
        target = targetSum;
        dfs(root);
        return result;
    }
};
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