Substring with Concatenation of All Words

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).

使用了 HashMap来降低时间复杂度，整个的算法很简单。

 1 public class Solution {
 2      int elen = 0;
 3     public ArrayList<Integer> findSubstring(String S, String[] L) {
 4         // Note: The Solution object is instantiated only once and is reused by each test case.
 5         ArrayList<Integer> result = new ArrayList<Integer>();
 6         if(S == null || S.length() == 0) return result;
 7         int slen = S.length();
 8         int n = L.length;
 9         elen = L[0].length();
10         HashMap<String, Integer> hm = new HashMap<String, Integer>();
11         for(int i = 0; i < n; i ++){
12             if(hm.containsKey(L[i])) 
13                 hm.put(L[i], hm.get(L[i]) + 1);
14             else                     
15                 hm.put(L[i], 1);
16         }
17         for(int i = 0; i <= slen - n * elen; i ++){
18             if(hm.containsKey(S.substring(i, i + elen)))
19                 if(checkOther(new HashMap<String, Integer>(hm), S, i))
20                     result.add(i);
21         }
22         return result;
23     }
24     public boolean checkOther(HashMap<String, Integer> hm, String s, int pos){
25         if(hm.size() == 0)  return true;
26         if(hm.containsKey(s.substring(pos, pos + elen))){
27             if(hm.get(s.substring(pos, pos + elen)) == 1)  
28                 hm.remove(s.substring(pos, pos + elen));
29             else                                           
30                 hm.put(s.substring(pos, pos + elen), hm.get(s.substring(pos, pos + elen)) - 1);
31             return checkOther(hm, s, pos + elen);
32         }
33         else return false;
34     }
35 }