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 }