Leetcode 395. Longest Substring with At Least K Repeating Characters 至少K重复最长子序列 解题报告
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1 解题思想
题目意思是说,给一个字符串和一个K,现在要找到一个子串,这个子串里面的每个出现的字符都至少出现K次。
这个不是一般的DP或者暴力的解决方式,要用一下递归:
1、统计每个字母【只有小写】出现的次数
2、找出不合法的字符【出现了,但是次数不够K】
3、如果没有任何不合法字符,那么返回串的长度
4、如果有不合法的字符,那么将这个串按照本轮不合法的字符位置切开(一个不合法的字符就切一刀),切开后的每一个子串递归执行1,取最长的返回
2 原题
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 = 3Output:3The longest substring is "aaa", as 'a' is repeated 3 times.Example 2: Input:s = "ababbc", k = 2Output:5The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.
3 AC解
public class Solution { public int longestSubstring(String s, int k) { //System.out.println(s); int n = s.length(); if(n < k) return 0; int counter[] = new int[26]; boolean valid[] = new boolean[26]; char ss[] = s.toCharArray(); //统计每个字符的长度 for(int i=0;i<n;i++) counter[ss[i] - 'a']++; //检查当前字符串是否是完全满足的 boolean fullValid = true; //判断每个字符的出现条件是否合适,即,要么不出现,要么出现了不少于k for(int i=0;i<26;i++){ if(counter[i]>0 && counter[i]<k){ valid[i] = false; fullValid = false; } else valid[i] = true; } if(fullValid) return s.length(); int max = 0; int lastStart=0; //把不符合要求的断开,然后依次检查 取最大 for(int i=0;i<n;i++){ if(valid[ss[i] - 'a'] == false){ // System.out.println(lastStart+" "+i); max = Math.max(max,longestSubstring(s.substring(lastStart,i),k)); lastStart = i + 1; } } max = Math.max(max,longestSubstring(s.substring(lastStart,n),k)); return max; }}
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