Substring with Concatenation of All Words

题目

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).

方法

使用map来存储。并进行推断。

    public List<Integer> findSubstring(String S, String[] L) {
        List<Integer> list = new ArrayList<Integer>();
        int arrLen = L.length;
        int strLen = L[0].length();
        Map<String, Integer> map = new HashMap<String, Integer>();
        for (int i = 0; i < arrLen; i++) {
        	if (map.get(L[i]) == null) {
        		map.put(L[i], 1);
        	} else {
        		map.put(L[i], map.get(L[i]) + 1);
        	}
        }
        
        Map<String, Integer> tempMap = new HashMap<String, Integer>();
        for (int i = 0; i < S.length() - arrLen * strLen + 1; i++) {
        	tempMap.clear();
        	boolean flag = true;
        	for (int j = 0; j < arrLen && flag; j++) {
        		String str = S.substring(i + j * strLen, i + (j + 1 )* strLen);
        		if (map.containsKey(str)) {
        			tempMap.put(str, tempMap.get(str) == null ? 1 : tempMap.get(str) + 1);
            		if (tempMap.get(str) > map.get(str)) {
            			flag = false;
            		}
        		} else {
        			flag = false;
        		}

        	}
        	if (flag) {
        		list.add(i);
        	}
        }
        return list;
    }