Substring with Concatenation of All Words

Problem:

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).

Analysis:

一开始觉得非常复杂，找了几个别人的博客看。看了几天，没看懂。最后找到一个牛人的地址http://n00tc0d3r.blogspot.com/2013/06/substring-with-concatenation-of-all.html?view=classic 慢慢的就懂了。不过有些细节还是不太清楚，只有慢慢积累。

现在开始把代码写到Eclipse里面了。1.可以节约在LeetCode编码的时间. 2. LeetCode如果报错可以Debug。之前一旦LeetCode没通过要找n久的bug，效率实在太低。

时间和空间复杂度就暂时不讨论了，不懂。

Solution:

1:  public class Solution {  
2:   public ArrayList<Integer> findSubstring(String S, String[] L){  
3:         ArrayList<Integer> res=new ArrayList<Integer>();  
4:         if(S==null||L==null)  
5:              return res;  
6:         HashMap<String,Integer> words=new HashMap<String,Integer>();  
7:         int wordLen=L[0].length(),total=L.length;  
8:         for(String s:L)//Store all words to hashmap  
9:         {  
10:              if(words.containsKey(s))  
11:                   words.put(s, words.get(s)+1);  
12:              else  
13:                   words.put(s, 1);  
14:         }  
15:         for(int i=0;i<=S.length()-wordLen*total;i++)  
16:         {  
17:              HashMap<String,Integer> curWords=new HashMap<String,Integer>(words);  
18:              for(int j=i;j<=S.length()-wordLen && !curWords.isEmpty();j+=wordLen)  
19:              {  
20:                   String word=S.substring(j,j+wordLen);  
21:                   if(!curWords.containsKey(word))  
22:                        break;  
23:                   else  
24:                   {  
25:                        if(curWords.get(word)>1)  
26:                             curWords.put(word, curWords.get(word)-1);  
27:                        else  
28:                             curWords.remove(word);  
29:                   }  
30:              }  
31:              if(curWords.isEmpty())  
32:                   res.add(i);  
33:         }  
34:       return res;  
35:    }  
36:  }