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。

public class Solution {
     int elen = 0;
    public ArrayList<Integer> findSubstring(String S, String[] L) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(S == null || S.length() == 0) return result;
        int slen = S.length();
        int n = L.length;
        elen = L[0].length();
        HashMap<String, Integer> hm = new HashMap<String, Integer>();
        for(int i = 0; i < n; i ++){
            if(hm.containsKey(L[i])) 
                hm.put(L[i], hm.get(L[i]) + 1);
            else                     
                hm.put(L[i], 1);
        }
        for(int i = 0; i <= slen - n * elen; i ++){
            if(hm.containsKey(S.substring(i, i + elen)))
                if(checkOther(new HashMap<String, Integer>(hm), S, i))
                    result.add(i);
        }
        return result;
    }
    public boolean checkOther(HashMap<String, Integer> hm, String s, int pos){
        if(hm.size() == 0)  return true;
        if(hm.containsKey(s.substring(pos, pos + elen))){
            if(hm.get(s.substring(pos, pos + elen)) == 1)  
                hm.remove(s.substring(pos, pos + elen));
            else                                           
                hm.put(s.substring(pos, pos + elen), hm.get(s.substring(pos, pos + elen)) - 1);
            return checkOther(hm, s, pos + elen);
        }
        else return false;
    }
}