438. Find All Anagrams in a String

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".
分析：
这题和mininum window几乎一摸一样，使用两个pointer来找这样的window。

public class Solution {
    public List<Integer> findAnagrams(String s, String t) {
        List<Integer> result = new LinkedList<>();
        if (t.length() > s.length()) {
            return result;
        }

        // get count of each letter
        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < t.length(); i++) {
            char ch = t.charAt(i);
            map.put(ch, map.getOrDefault(ch, 0) + 1);
        }
        
        int begin = 0, end = 0;
        int counter = t.length();
        while (end < s.length()) {
            char endChar = s.charAt(end);
            if (map.containsKey(endChar)) {
                if (map.get(endChar) > 0) {
                    counter--;
                }
                map.put(endChar, map.get(endChar) - 1);
            }
            end++;

            while (counter == 0) {
                if (end - begin == t.length()) {
                    result.add(begin);
                }
                char beginChar = s.charAt(begin);
                if (map.containsKey(beginChar)) {
                    map.put(beginChar, map.get(beginChar) + 1);
                    if (map.get(beginChar) > 0) {
                        counter++;
                    }
                }
                begin++;
            }
        }
        return result;
    }
}