[LeetCode] (medium) 424. Longest Repeating Character Replacement

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博客围绕LeetCode上的题目展开，给定仅含大写字母的字符串，可最多k次替换字母，求最长重复字母子串长度。还提及滑动窗口方法，并指出其写法有技巧性，给出标准写法链接。

https://leetcode.com/problems/longest-repeating-character-replacement/

Given a string that consists of only uppercase English letters, you can replace any letter in the string with another letter at most k times. Find the length of a longest substring containing all repeating letters you can get after performing the above operations.

Note:
Both the string's length and k will not exceed 104.

Example 1:
Input:s = "ABAB", k = 2
Output:4

Explanation:
Replace the two 'A's with two 'B's or vice versa.
Example 2:
Input:s = "AABABBA", k = 1
Output:4

Explanation:
Replace the one 'A' in the middle with 'B' and form "AABBBBA".
The substring "BBBB" has the longest repeating letters, which is 4.

https://www.cnblogs.com/reboot329/p/5968393.html

滑动窗口：

class Solution {
public:
    int characterReplacement(string s, int k) {
        if(s.size() < k+2) return s.size();
        
        int lo = 0;
        int hi = k;
        int result = k+1;
        char maj = 'A';
        vector<int> cnt(26, 0);
        
        for(int i = lo; i <= hi; ++i){
            ++cnt[s[i]-'A'];
        }

        for(int i = 0; i < 26; ++i){
            if(cnt[i] > cnt[maj-'A']) maj = 'A'+i;
        }
        
        ++hi;
        
        while(hi < s.size()){
            ++cnt[s[hi]-'A'];
            if(cnt[s[hi]-'A'] > cnt[maj-'A']) maj = s[hi];
            
            while(hi-lo+1-cnt[maj-'A'] > k){
                --cnt[s[lo]-'A'];
                if(s[lo] == maj){
                    for(int i = 0; i < 26; ++i){
                        if(cnt[i] > cnt[maj-'A']) maj = 'A'+i;
                    }
                }
                ++lo;
            }
            
            result = max(result, hi-lo+1);
            ++hi;
        }
        
        return result;
    }
};

滑动窗口的写法本身就很有技巧性，贴一个标准写法：

class Solution {
public:
    int characterReplacement(string s, int k) {
        int n = s.length();
        if (n == 0) return 0;
        int cnt[26] = {0};
        int maxCount = 0, prev = 0;
        for (int i = 0; i < n; i++) {
            cnt[s[i]-'A']++;
            maxCount = max(maxCount, cnt[s[i]-'A']);
            if (maxCount+k < i-prev+1) {
                cnt[s[prev]-'A']--;
                prev++;
            }
        }
        return n-prev;
    }
};