Substring with Concatenation of All Words

原题：

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).

解答：

public List<Integer> findSubstring(String S, String[] L) {
		ArrayList<Integer> result = new ArrayList<Integer>();
		if(S==null || S.isEmpty() || L==null || L.length <1) {
			return result;
		}
		HashMap<String, Integer> toFind = new HashMap<String, Integer>();
		HashMap<String, Integer> find = new HashMap<String, Integer>();
		int arr_len = L.length;
		int str_len = L[0].length();
		for(int i=0; i<arr_len; i++) {
			if(!toFind.containsKey(L[i])) {
				toFind.put(L[i], 1);
			} else {
				toFind.put(L[i], toFind.get(L[i])+1);
			}
		}
		for(int i=0; i<=S.length()-arr_len*str_len; i++) {
			find.clear();
			int j;
			for(j=0; j<arr_len; j++) {
				int k = i+j*str_len;
				String subStr = S.substring(k, k+str_len);
				if(!toFind.containsKey(subStr)) break;
				if(!find.containsKey(subStr)) {
					find.put(subStr, 1);
				} else {
					find.put(subStr, find.get(subStr)+1);
				}
				if(find.get(subStr) > toFind.get(subStr)) break;
			}
			if(j == arr_len) result.add(i);
		}
		return result;
    }