30. Substring with Concatenation of All Words

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You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: “barfoothefoobarman”
words: [“foo”, “bar”]

You should return the indices: [0,9].
(order does not matter).

解法1： HashMap ＋　一次遍历　

用HashMap记录每一个word出现的次数，然后对每一个起始点进行遍历。
复杂度应该是O(NL), N是字符串的长度，L是字符串数组的个数。

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class Solution {
    public static List<Integer> findSubstring(String S, String[] L) {
        List<Integer> res = new ArrayList<Integer>();
        if (S == null || L == null || L.length == 0) return res;
        int len = L[0].length(); // length of each word
        Map<String, Integer> map = new HashMap<String, Integer>(); // map for L
        for (String w : L) map.put(w, map.containsKey(w) ? map.get(w) + 1 : 1);
        for (int i = 0; i <= S.length() - len * L.length; i++) {
            Map<String, Integer> copy = new HashMap<String, Integer>(map);
            for (int j = 0; j < L.length; j++) { // checkc if match
                String str = S.substring(i + j*len, i + j*len + len); // next word
                if (copy.containsKey(str)) { // is in remaining words
                    int count = copy.get(str);
                    if (count == 1) copy.remove(str);
                    else copy.put(str, count - 1);
                    if (copy.isEmpty()) { // matches
                        res.add(i);
                        break;
                    }
                } else break; // not in L
            }
        }
        return res;
    }
}