[LeetCode] 395. Longest Substring with At Least K Repeating Characters

题：https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/description/

题目

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.

思路

题目大意

一个字符串 Str，都是由 小写字母组成。找出一个最长子串 subStr，这子串包含的 字符在子串中出现的次数大于等于 k。

解题思路

总体思路将 整体求解 分解为 局部求解。

1. 对于 字符串 s。统计字符串 s 中字符出现次数 小于 k的字符，将字符放入 fliterSet 集合中。
2. 若 fliterSet 中元素为空，那么 s 中 符合要求的 最大子串 就是 s本身，长度为 len(s)。否则，用 fliterSet 的元素将 s分割，将切割的多个子字符串strLst。将strLst中的每个 字符串 进行 1中的处理，获得 最长子串数。

Note：
代码 1 比较 简洁，代码 2 便于 理解。都是实现上面的思路。

code

1

class Solution:
    def longestSubstring(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        if len(s) == 0:
            return 0
        counter = {}

        for i in range(len(s)):
            if s[i] in counter.keys():
                counter[s[i]] += 1
            else:
                counter[s[i]] = 1
        filter_set = set()

        for key in counter.keys():
            if counter[key] < k:
                filter_set.add(key)
        if len(filter_set) == 0:
            return len(s)
        subsmaxlen = 0
        starti = 0
        i = 0
        while i < len(s):
            if s[i] in filter_set:
                if starti != i:
                    subsmaxlen = max(subsmaxlen, self.longestSubstring(s[starti:i], k))
                starti = i+1
            i += 1
        if i - starti >0:
            subsmaxlen = max(subsmaxlen, self.longestSubstring(s[starti:i], k))

        return subsmaxlen

2

class Solution:
    def longestSubstring(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        counter = {}

        for i in range(len(s)):
            if s[i] in counter.keys():
                counter[s[i]] += 1
            else:
                counter[s[i]] = 1
        filter_set = set()

        for key in counter.keys():
            if counter[key] < k:
                filter_set.add(key)
        if len(filter_set) == 0:
            return len(s)
        slst = list(s)
        for i in range(len(slst)):
            if slst[i] in filter_set:
                slst[i] = '#'

        subsmaxlen = 0
        substring = "".join(slst).split("#")
        for i in range(len(substring)):
            subsmaxlen = max(subsmaxlen, self.longestSubstring(substring[i], k))
        return subsmaxlen