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".

 1 public class Solution {
 2     public List<Integer> findAnagrams(String s, String p) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         if (s==null || s.length()==0 || p==null || p.length()==0 || p.length()>s.length()) return res;
 5         int[] chars = new int[26];
 6         for (int i=0; i<p.length(); i++) {
 7             chars[p.charAt(i)-‘a‘]++;
 8         }
 9         int start = 0, end = 0, count = p.length();
10         while (end < s.length()) {
11             if (end-start==p.length() && chars[s.charAt(start++)-‘a‘]++>=0) count++;
12             if (--chars[s.charAt(end++)-‘a‘] >= 0) count--;
13             if (count == 0) res.add(start);
14         }
15         return res;
16     }
17 }