11-23 每日一题：859. 亲密字符串

Leetcode 859. 亲密字符串

思路：统计每个字符出现的频数

亲密字符串必须保证每个字符出现的频数一样多，判断不相同的位置即可

时间复杂度：$O(n)$

空间复杂度：$O(C)$

class Solution {
public:
    bool buddyStrings(string s, string goal) {
        int n = s.size(), m = goal.size();
        if (n != m) return false;
        char a[26] = {0}, b[26] = {0};
        int cnt = 0;
        for (int i = 0; i < n; i ++ ) {
            if (s[i] != goal[i]) 
                cnt ++ ;
            a[s[i] - 'a'] ++ ;
            b[goal[i] - 'a'] ++ ;
        }
        bool flag = false;
        for (int i = 0; i < 26; i ++ ) {
            if (a[i] != b[i]) return false;
            if (a[i] > 1) flag = true;
        }
        return cnt == 2 || (!cnt && flag);
    }
};