Leetcode Substring with Concatenation of All Words

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].

Difficulty: Hard

Solution: 

HashMap<String, Queue<Integer>>

public class Solution {    List<Integer> res = new ArrayList<Integer>();    String[] wordList;    HashMap<String, Queue<Integer>> map = new HashMap<String, Queue<Integer>>();        public List<Integer> findSubstring(String s, String[] words) {        if(words.length == 0) return res;        int len = words[0].length();                for(int i = 0; i < words.length; i++){            if(map.containsKey(words[i])){                map.get(words[i]).offer(-1);            }            else{               Queue<Integer> q = new LinkedList<Integer>();               q.offer(-1);               map.put(words[i], q);            }        }        wordList = new String[s.length() - len + 1];        for(int i = 0; i < wordList.length; i++){            wordList[i] = s.substring(i, i + len);        }        for(int i = 0; i < len; i++){            int begin = 0, index = i, count = 0;            while(begin < s.length()){                // System.out.println(index);                // System.out.println(index - len * count);                // System.out.println("---");                if(words.length == count){                    //System.out.println("ADD");                    res.add(index - len * count);                    count--;                    continue;                }                if((words.length - count) * len > s.length() - index){                    break;                }                String newWord = wordList[index];                if(!map.containsKey(newWord)){                    index += len;                    count = 0;                    continue;                }                if(map.get(newWord).peek() >= index - count * len && map.get(newWord).peek() < index && (index - map.get(newWord).peek()) % len == 0){                    int temp = map.get(newWord).poll();                    map.get(newWord).offer(index);                    count = (index - temp)/len;                    index += len;                    continue;                }                else{                    map.get(newWord).poll();                    map.get(newWord).offer(index);                    index += len;                    count++;                }            }            if(words.length == count)                res.add(index - len * count);        }        return res;    }}