题目

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).

思路

这么说吧。我是通过比较作的。所以为了减少寻找时间，用了HashMap来做，其key是L中对应的字符串，其Value是该字符串出现的次数

然后通过2个指针来做

因为L的每个长度一样，所以我们可以对S进行截断比较，如果找到，就把对应的temp中的值减一或者是直接删除掉（这里需要一个temp来保存HashMap中的数据，每进行一次循环比较，就重新赋值一次。原因就是要删除temp中的数据，保证答案的正确性）

最后如果次数达到了L的长度，就表示找到了.

但是我用这个代码提交，有的时候会报超时，有的时候却AC。。无语..

估计是哪里写不好...还要努力

代码

public class Solution {
    	 public ArrayList<Integer> findSubstring(String S, String[] L)
	 {
		 
		 ArrayList<Integer> result = new ArrayList<Integer>();
		 HashMap<String,Integer> map = new HashMap<String, Integer>();
		 if(L.length == 0)
			 return result;
	 
		 for(int i = 0 ; i<L.length; i ++)
		 {			 
			 if(map.containsKey(L[i]))
			 {
				 map.put(L[i], map.get(L[i])+1);
			 }
			 else
			 {
				 map.put(L[i], 1);
			 }
		 }	 	 
		 	
		 int sLength = S.length();
		 int eachLength = L[0].length();
		 int numL = L.length;
		 
		 for(int i =0 ; i <= sLength-eachLength*numL;i++)
		 {
			 boolean tag = false;
			 int times = 0;
			
			 for(int j = i; j < i + eachLength*numL;j+=eachLength)
			 {
				String sub =  S.substring(j, j+eachLength);
				
				if (map.containsKey(sub))
				{
					
					if(map.get(sub)>1)
					{
						map.put(sub, map.get(sub)-1);
					}
					
					times++;
					
					
					if(times == numL)
					{
						tag = true;
					}
				}
				
			 }
			 if(tag == true)
			 {
				result.add(i); 
			 }
		 }
		 return result;


	 }
}

LeetCode|Substring with Concatenation of All Words,布布扣,bubuko.com

LeetCode|Substring with Concatenation of All Words

原文：http://blog.csdn.net/hwb1992/article/details/24364625