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：用一个hashmap保存链表L，对S进行扫描，若每个子单词在L中出现且仅出现一次，则满足条件。

public class Solution {
    public ArrayList<Integer> findSubstring(String S, String[] L) {
        ArrayList<Integer> ret = new ArrayList<Integer>();
    	if (S == null || L == null) {
			return ret;
		}
    	int length_words = L[0].length();
    	int length_list = L.length;
    	int length_str = length_words*length_list;
    	
    	HashMap<String, Integer> list_words = new HashMap<String, Integer>();
    	for (String word : L) {
			if (!list_words.containsKey(word)) {
				list_words.put(word, 1);
			}
			else list_words.put(word, list_words.get(word)+1);
		}
    	
    	for (int i = 0; i <= S.length()-length_str; i++) {
			String tmp = S.substring(i, i+length_str);
			
			
			HashMap<String, Integer> tmp_hashHashMap = (HashMap<String, Integer>)list_words.clone();
			int j;
			for (j = 0; j < length_list; j++) {
				String tmpword = tmp.substring(j*length_words, j*length_words+length_words);
				if (!tmp_hashHashMap.containsKey(tmpword)) {
					break;
				}
				else if (tmp_hashHashMap.get(tmpword) <= 0) {
					break;
				}
				else tmp_hashHashMap.put(tmpword, tmp_hashHashMap.get(tmpword)-1);
			}
			
			if (j == length_list) {
				ret.add(i);
			}
		}
    	return ret;
    }
}