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".

Very Important:

Sliding Window algorithm template to solve all the Leetcode substring search problem.
link

Using slidewindow to solve this

Source

My version

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<Integer>();
        int[] arr = new int[26];
        char[] ch = p.toCharArray();
        char[] chs = s.toCharArray();

        for(char c:ch) arr[c-'a']++;

        for(int i=0;i<=chs.length-ch.length;i++){
            if(arr[chs[i]-'a']==0) continue;
            int cnt = 0;
            int j=i;
            for(;j<i+ch.length;j++){
                if(arr[chs[j]-'a']==0) {
                    break;
                }else {
                    arr[chs[j]-'a']--;
                    cnt++;
                }
            }
            if(cnt==ch.length) res.add(i);
            for(int m=i;m<j;m++) arr[chs[m]-'a']++;
        }
        return res;
    }
}