LeetCode賽題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

先統計出字符串中每個字符出現的次數,對於出現次數少於k的字符,任何一個包含該字符的字符串都不是符合要求的子串,因此這樣的字符就是分隔符,應該以這些出現次數少於k次的字符做分隔符打斷原字符串,然后對各個打斷得到的字符串進行遞歸統計,得到最長的符合要求的字符串。如果一個字符串中不包含分隔符(即每個字符出現的次數都達到了k次及以上次數),那么這個字符串就是符合要求的子串。

Java 算法實現:

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){
        	//說明有分隔符
        	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{
        	//沒有分隔符,說明s中的每一個字符出現的次數都大於等於k
        	return s.length();
        }
    }
}


免責聲明!

本站轉載的文章為個人學習借鑒使用,本站對版權不負任何法律責任。如果侵犯了您的隱私權益,請聯系本站郵箱yoyou2525@163.com刪除。



 
粵ICP備18138465號   © 2018-2025 CODEPRJ.COM