面试题 01.02. 判定是否互为字符重排

面试题 01.02. 判定是否互为字符重排 - 力扣（LeetCode）

简单的哈希表思路，遍历计数即可：

class Solution {
public:
    bool CheckPermutation(string s1, string s2) {
        if(s1.size() != s2.size())  return false;
        int count[26] = {0};
        for(int i = 0; i < s1.size(); ++i){
            ++count[s1[i]-'a'];
            --count[s2[i]-'a'];
        }
        for(int i = 0; i < 26; ++i){
            if(count[i] != 0)   return false;
        }
        return true;
    }
};

或者排序：

class Solution {
public:
    bool CheckPermutation(string s1, string s2) {
        if(s1.size() != s2.size())  return false;
        sort(s1.begin(), s1.end());
        sort(s2.begin(), s2.end());
        return s1 == s2;
    }
};