[Leetcode] 424. Longest Repeating Character Replacement 解题报告

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本文介绍了一种使用滑动窗口算法解决字符串中字符替换问题的方法，旨在寻找经过k次替换后形成的最长重复字符子串。

题目：

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.

思路：

这道题目最简单的思路就是用移动窗口来解决。对于一个长度为m的子串而言，我们假设里面出现次数最多的字符的出现次数是n，那么m可以在k次之内被替换成为单一字符的充要条件是m - n <= k。因此，我们仅需要维护一个滑动窗口，使得这个窗口的长度减去出现次数最多的字符的出现次数小于等于k。在遍历的过程中找到这个最长的m即可。

我们下面实现的算法的空间复杂度是O(1)，时间复杂度是O(n)，这里n是输入字符串的长度。为什么for循环里面的while循环不额外增加时间复杂度呢？这是因为每个字符在while循环内最多被去除一次，因此while总的循环次数不会超过O(n)。

代码：

class Solution {
public:
    int characterReplacement(string s, int k) {
        int size = s.size();
        vector<int> count(26,0);
        int begin = 0, maxlen = 0, ans = 0;
        for(int end = 0; end < size; ++end) {
            maxlen = max(maxlen, ++count[s[end] - 'A']);    // s[end] is the new possible dominate character
            while(end - begin + 1 - maxlen > k) {           // length(substring) - maxlen > k mean we cannot form repeating letters
                --count[s[begin++] - 'A'];
            }
            ans = max(ans,  end - begin + 1);
        }
        return ans;
    }
};