Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:s: "cbaebabacd" p: "abc"Output:[0, 6]Explanation:The substring with start index = 0 is "cba", which is an anagram of "abc".The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:s: "abab" p: "ab"Output:[0, 1, 2]Explanation:The substring with start index = 0 is "ab", which is an anagram of "ab".The substring with start index = 1 is "ba", which is an anagram of "ab".The substring with start index = 2 is "ab", which is an anagram of "ab".

public static List
    
      findAnagrams(String s, String p) {        List
     
       list = new ArrayList<>();        if(s.length() < p.length()){            return list;        }        for(int i = 0;i<=(s.length()-p.length());i++){            if(isAnagram(i,s,p)){                list.add(i);            }        }        return list;    }    private static boolean isAnagram(int begin, String s, String p) {        //这样子还是不能去重//        List
      
        list = new ArrayList<>();//        for(int j= 0;j