LeetCode赛题395----Longest Substring with At Least K Repeating Characters

395. Longest Substring with At least K Repeating Characters

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.

算法分析

先统计出字符串中每个字符出现的次数，对于出现次数少于k的字符，任何一个包含该字符的字符串都不是符合要求的子串，因此这样的字符就是分隔符，应该以这些出现次数少于k次的字符做分隔符打断原字符串，然后对各个打断得到的字符串进行递归统计，得到最长的符合要求的字符串。如果一个字符串中不包含分隔符(即每个字符出现的次数都达到了k次及以上次数)，那么这个字符串就是符合要求的子串。

Java 算法实现：

public class Solution {
    public int longestSubstring(String s, int k) {
        if(k<=1){
        	return s.length();
        }

        int[] repeat=new int['z'+1];
        for(int i=0;i<s.length();i++){
        	repeat[s.charAt(i)]++;
        }
        StringBuilder regex=new StringBuilder("");
        boolean firstSplit=true;
        for(int i='a';i<='z';i++){
        	if(repeat[i]>0&&repeat[i]<k){
        		if(firstSplit){
        			regex.append((char)i);
        			firstSplit=false;
        		}
        		else{
        			regex.append("|"+(char)i);
        		}
        	}
        }
        if(regex.length()>0){
        	//说明有分隔符
        	String[] strs=s.split(regex.toString());
        	int max=0;
        	int tmpAns=0;
        	for(String str:strs){
        		tmpAns=longestSubstring(str, k);
        		if(tmpAns>max){
        			max=tmpAns;
        		}
        	}
        	return max;
        }
        else{
        	//没有分隔符,说明s中的每一个字符出现的次数都大于等于k
        	return s.length();
        }
    }
}