LeetCode每日一题：substring with concatenation of all words

问题描述

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).

问题分析

这题其实就是子串匹配，每次截取部分子串来与字典进行匹配即可，由于字典中长度是确定的，所以很方便进行循环，最外循环只需要O(n)即可。思路是对的，但是老是不能AC，最后只能从网上抄了一个解法...
详情见CSDN博客，最后要把res排下序，要不过不了AC。

代码实现

public ArrayList findSubstring(String S, String[] L) {
        ArrayList res = new ArrayList();
        if (S == null || S.length() == 0 || L == null || L.length == 0)
            return res;
        HashMap map = new HashMap();
        for (int i = 0; i < L.length; i++) {
            if (map.containsKey(L[i])) {
                map.put(L[i], map.get(L[i]) + 1);
            } else {
                map.put(L[i], 1);
            }
        }
        for (int i = 0; i < L[0].length(); i++) {
            HashMap curMap = new HashMap();
            int count = 0;
            int left = i;
            for (int j = i; j <= S.length() - L[0].length(); j += L[0].length()) {
                String str = S.substring(j, j + L[0].length());

                if (map.containsKey(str)) {
                    if (curMap.containsKey(str))
                        curMap.put(str, curMap.get(str) + 1);
                    else
                        curMap.put(str, 1);
                    if (curMap.get(str) <= map.get(str))
                        count++;
                    else {
                        while (curMap.get(str) > map.get(str)) {
                            String temp = S.substring(left, left + L[0].length());
                            if (curMap.containsKey(temp)) {
                                curMap.put(temp, curMap.get(temp) - 1);
                                if (curMap.get(temp) < map.get(temp))
                                    count--;
                            }
                            left += L[0].length();
                        }
                    }
                    if (count == L.length) {
                        res.add(left);
                        String temp = S.substring(left, left + L[0].length());
                        if (curMap.containsKey(temp))
                            curMap.put(temp, curMap.get(temp) - 1);
                        count--;
                        left += L[0].length();
                    }
                } else {
                    curMap.clear();
                    count = 0;
                    left = j + L[0].length();
                }
            }
        }
        Collections.sort(res);
        return res;
    }