Longest Repeating Character Repl

题目来源
给一个字符串和一个整数k，可以变化k个字符，使得字符串中连续重复的字符数最多。
我想着一个移动窗口，里面字符个数减去频次最高的字符个数等于k，然后从前往后移动，计算窗口最大的时候窗口内元素个数即为所求。
代码如下：

class Solution {
public:
    int characterReplacement(string s, int k) {
        int n = s.size();
        if (n <= k + 1)
            return n;
        vector<int> chaNum(26, 0);
        int longest = k;
        int l = 0, r = k;
        for (int i=0; i<=k; i++)
            chaNum[s[i]-'A']++;
        while (r < n) {
            int most = 0;
            for (int i=0; i<26; i++)
                most = max(most, chaNum[i]);
            if (r - l + 1 - most <= k) {
                longest = max(longest, r-l+1);
                r++;
                chaNum[s[r]-'A']++;
            }
            else {
                chaNum[s[l]-'A']--;
                l++;
            }
        }
        return longest;
    }
};

稍作修改，看着简洁一些。

class Solution {
public:
    int characterReplacement(string s, int k) {
        int n = s.size();
        if (n <= k + 1)
            return n;
        vector<int> chaNum(26, 0);
        int longest = k + 1;
        int l = 0, r = 0;
        while (r < n) {
            chaNum[s[r]-'A']++;
            int most = *max_element(chaNum.begin(), chaNum.end());
            while (r - l + 1 - most > k)
                chaNum[s[l++]-'A']--;
            longest = max(longest, r-l+1);
            r++;
        }
        return longest;
    }
};