Leetcode 395. Longest Substring with At Least K Repeating Characters-优快云博客

本文链接：https://blog.youkuaiyun.com/sundawei2016/article/details/76599422

本文介绍了一种寻找字符串中每个字符至少重复k次的最长子串的算法。通过递归方式，利用辅助函数筛选出不符合条件的字符并分割字符串，最终求得最长符合条件的子串长度。

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Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times.

Example 1:

Input:
s = "aaabb", k = 3

Output:
3

The longest substring is "aaa", as 'a' is repeated 3 times.

Example 2:

Input:
s = "ababbc", k = 2

Output:
5

The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.

找出小于 K 的字符串。

如果没有小于 K 的字符串，则返回字符串 s 的长度

然后使用归递来找最大的长度

    public int longestSubstring(String s, int k) {
        if (s.length() < k) return 0;
        String lessThanK = helper(s, k);
        if (lessThanK.length() == 0) return s.length();
        String[] tokens = s.split("[" + lessThanK + "]");
        int maxLen = 0;
        for (String token : tokens) {
            maxLen = Math.max(maxLen, longestSubstring(token, k));
        }
        return maxLen;
    }
    
    private String helper(String s, int k) {
        int[] map = new int[26];
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            map[c - 'a']++;
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < map.length; i++) {
            if (map[i] > 0 && map[i] < k) sb.append((char) (i + 'a'));
        }
        return sb.toString();
    }