30. Substring with Concatenation of All Words

感觉有点像暴力循环破解法，虽然能accepted,但有点慢

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {

        Set<String> number = new HashSet<>();
        List<Integer> list = new ArrayList<>();
        int len = words[0].length();//有几个字符串
        int allLen = words.length*len;//words中所有字符串加起来的长度
        String keyWord="";
        HashMap<String,Integer> map = new HashMap<>();
        for (String word:words) {
                map.put(word,map.getOrDefault(word,0)+1);
            }
        for (String word:words) {
          number.add(word);
        }
//枚举法开始，从开头到结尾枚举从start字段到（start➕allLen）下标字段(这部分刚刚好是allLen长度的字符串)是不是符合情况，这个逻辑蛮简单。
        for (int i =0 ;  i<s.length()-allLen+1;i++){

            HashMap<String,Integer> hashMap = new HashMap(map);

            int cc = number.size();
            for (int start=i; start<i+allLen;start+=len){
                keyWord = s.substring(start,start+len);
                if(hashMap.containsKey(keyWord)){
                    hashMap.put(keyWord,hashMap.get(keyWord)-1);
                    if (hashMap.get(keyWord)==0){
                        cc--;
                    }else if (hashMap.get(keyWord)<0){
                        break;
                    }else {
                        continue;
                    }
                }
                else break;
            }

            if (cc==0)list.add(i);

        }

        return list;
    }
}