30. 串联所有单词的子串 Substring with Concatenation of All Words

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

You can return the answer in any order.

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.

方法：

注意words数组的长度都是一样的，这样可以简化查找

将words里的元素放到map里，s按照word长度截取，放到另外一个map

两个map相等则返回起始位置

public List<Integer> findSubstring(String s, String[] words) {
        List<Integer>  ans = new ArrayList<>();
        if(s == null || words == null || words.length == 0){
            return ans;
        }
        int len = words[0].length();
        Map<String,Integer> map = new HashMap<>();
        for (String word : words){
            if (map.containsKey(word)){
                map.put(word, map.get(word) + 1);
            }else{
                map.put(word, 1);
            }
        }
        for (int  i = 0; i < s.length() - words.length * len + 1; i++){
            Map<String,Integer> map_s = new HashMap<>();
            for (int  j = 0; j < words.length * len; j += len){
                if (!map.containsKey(s.substring(i+j,i+j+len))) break;
                if (map_s.containsKey(s.substring(i+j,i+j+len))){
                    map_s.put(s.substring(i+j,i+j+len), map_s.get(s.substring(i+j,i+j+len)) + 1);
                }else{
                    map_s.put(s.substring(i+j,i+j+len), 1);
                }
            }
            if (map.equals(map_s)) ans.add(i);
        }


        return ans;

    }