[leetcode]-30-Substring with Concatenation of All Words

题目：

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.

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:
  s = "wordgoodgoodgoodbestword",
  words = ["word","good","best","word"]
Output: []

思路：

遍历字符串s，每次取出和words一样长的字串，分割成数组subs，比较数组subs和words，若相同（不考虑顺序），则找到一个index放入结果集中。

代码：

     public List findSubstring(String s, String[] words) {
        List res = new LinkedList<>();
        if(words == null || words.length == 0) return res;
        int wordLength = words[0].length();
        for(int i = 0; i <= s.length() - wordLength*words.length; ++i){
            String[] subs = new String[words.length];
            int k = i;
            for(int j = 0; j < words.length; ++j) {
                subs[j] = s.substring(k, k + wordLength);
                k += wordLength;
            }
            if(compareTwoArrays(subs, words)){
                res.add(i);
            }
        }
        return res;
    }
    public boolean compareTwoArrays(String[] subs, String[] words){
        HashMap map = new HashMap<>();
        for(int i = 0; i < words.length; ++i){
            if(map.getOrDefault(words[i], 0) == 0){
                map.put(words[i],1);
            }else{
                map.put(words[i], map.get(words[i])+1);
            }
        }
        for(int i = 0; i < subs.length; ++i){
            if(map.getOrDefault(subs[i],0) == 0)
                return false;
            else
                map.put(subs[i], map.get(subs[i])-1);
        }
        for(int i = 0; i < words.length; ++i){
            if(map.getOrDefault(words[i], 0) != 0){
                return false;
            }
        }
        return true;
    }