leetcode JAVA Substring with Concatenation of All Words 难度系数3 3.12_ou are given a string, s, and a list of words, wor-优快云博客

本文介绍了一种高效的字符串匹配算法，该算法能在给定的字符串中找到由多个指定长度相同的单词组成的子串的所有起始位置。通过使用哈希表来跟踪每个单词出现的次数，并遍历字符串检查所有可能的组合，实现了快速定位所有符合条件的子串。

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Question：

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

import java.util.ArrayList;
import java.util.Hashtable;
public class Solution {
    public ArrayList<Integer> findSubstring(String S, String[] L) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        Hashtable<String,Integer> ht = new Hashtable<String, Integer>();
        for(String str: L){
        	if(ht.containsKey(str)){
        		ht.put(str,ht.get(str)+1);
        	}else {
				ht.put(str, 1);
			}
        }
        int wordlen = L[0].length();
        int wordnumber = L.length;
        int totallen = wordlen*wordnumber;
        for(int i =0;i<=S.length()-totallen;i++){
        	Hashtable<String,Integer> temp = new Hashtable<>(ht);
        	int count = 0;
        	for(int j=i;j<=i+totallen-wordlen;j+=wordlen){
        		String sub = S.substring(j, j+wordlen);
        		if(temp.containsKey(sub)){
        			int times = temp.get(sub);
        			if (times==1) {
						temp.remove(sub);
					}else {
						temp.put(sub, times-1);
					}
        			count++;
        		}else {
					break;
				}
        	}
        	if(count==wordnumber){
        		res.add(i);
        	}
        }
        return res;
    }
}