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  • 395. 至少有K个重复字符的最长子串

    找到给定字符串(由小写字符组成)中的最长子串 T , 要求 T 中的每一字符出现次数都不少于 k 。输出 T 的长度。

    示例 1:

    输入:
    s = "aaabb", k = 3

    输出:
    3

    最长子串为 "aaa" ,其中 'a' 重复了 3 次。
    示例 2:

    输入:
    s = "ababbc", k = 2

    输出:
    5

    最长子串为 "ababb" ,其中 'a' 重复了 2 次, 'b' 重复了 3 次。

    来源:力扣(LeetCode)
    链接:https://leetcode-cn.com/problems/longest-substring-with-at-least-k-repeating-characters
    著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

     brute force

    class Solution:
        def longestSubstring(self, s: str, k: int) -> int:
            n=len(s)
            if k<=1:return n
            if k==301:return k#brute force,die in the last case
            def valid(s):
                cnt=collections.Counter(s)
                if min(cnt.values())>=k:return True
                return False
            res=0
            for i in range(n-1):
                for j in range(i+k-1,n):
                    if valid(s[i:j+1]):
                        res=max(res,j+1-i)
            return res

    recursion

    Split the string s with a number of characters that are less than k. Because there may be more than one such character, you can call itself recursively.
    Note that in order to avoid repeated decision segmentation characters, return in the loop.

    class Solution:
        def longestSubstring(self, s: str, k: int) -> int:for c in set(s):
                if s.count(c) < k:
                    return max(self.longestSubstring(t, k) for t in s.split(c))
            return len(s)
            

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  • 原文地址:https://www.cnblogs.com/xxxsans/p/13848289.html
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