Leetcode (OK)242 1 219 （思路）383 202 （*）205 290 49

383. Ransom Note(想一下思路)

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int record[26]{0};
 
        for(auto num:magazine){
            record[num-'a']++;
        }
 
        for(auto note:ransomNote){
            record[note-'a']--;
        }
 
        for(int i=0; i<26; i++){
            if(record[i] < 0)
                return false;
        }
 
        return true;
    }
};

205. Isomorphic Strings(*)

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        unordered_map<char, char> mp, mp2;
        for (int i=0; ilength(); ++i) {
            if(mp[s[i]] && mp[s[i]] != t[i]) return false;
            if(mp2[t[i]] && mp2[t[i]] != s[i]) return false;
            mp[s[i]] = t[i];
            mp2[t[i]] = s[i];
        }
        return true;
    }
};

290. Word Pattern(*)

class Solution {
public:
    bool wordPattern(string pattern, string s) {
        vector words;
        string sub_s;
 
        for(auto val:s){
            if(val == ' '){
                words.push_back(sub_s);
                sub_s = "";
            }else{
                sub_s.push_back(val);
            }
        }
        words.push_back(sub_s);
        if(pattern.size() != words.size()) return false;
 
        unordered_map<char, string> patternToWord;
        unordered_set wordSet;
 
        for(int i=0; isize(); ++i){
            string s = words[i];
            char ch = pattern[i];
            if(!patternToWord.count(ch)){
                if(wordSet.count(s))    return false;
                patternToWord[ch] = s;
            }else{
                if(patternToWord[ch] != s) return false;
            }
            wordSet.insert(s);
        }
 
        return true;
    }
};

1.注意如何存储单词

2.要先判断两个大小是否一样 

242. Valid Anagram

class Solution {
public:
    bool isAnagram(string s, string t) {
        int record[26]{0};
        for(auto ch:s){
            record[ch-'a']++;
        }
        for(auto ch:t){
            record[ch-'a']--;
        }
        for(int i=0; i<26; i++){
            if(record[i] != 0) return false;
        }
        return true;
    }
};

49. Group Anagrams(*)

class Solution {
private:
    string encode(string str){
        vector<int> count{26, 0};
        for(char ch:str){
            count[ch-'a'] ++;
        }
        string code(count.begin(), count.end());
        return code;
    }
public:
    vector> groupAnagrams(vector& strs) {
        unordered_map> codeToGroup;
        for(string s:strs){
            string code = encode(s);
            codeToGroup[code].push_back(s);
        }
 
        vector> res;
        for(auto group:codeToGroup){
            res.push_back(group.second);
        }
 
        return res;
    }
};

也可以直接用sort

1. Two Sum

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> record;
        for(int i=0; isize(); i++){
            int need = target-nums[i];
            if(record.count(need)){
                return {record[need], i};
            }
            record.emplace(nums[i], i);
        }
 
        return {};
    }
 
};

202. Happy Number(想一下思路)

class Solution {
private:
    int getSum(int n){
        int sum = 0;
        while(n >= 1){
            sum += (n%10) * (n%10);
            n /= 10; 
        }
        return sum;
    }
public:
    bool isHappy(int n) {
        unordered_set<int> record;
        while(1){
            n= getSum(n);
            if(n == 1) return true;
            if(record.count(n)) return false;
 
            record.insert(n);
 
        }
    }
};

219. Contains Duplicate II(OK)

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        unordered_map<int, int> record;
        for(int i=0; isize(); i++){
            if(record.count(nums[i])){
                int dif = i-record[nums[i]];
                if(dif <= k) return true;
                
            }
            record[nums[i]] = i;
        }
        return false;
    }
};