1、前言

2、思路

这道题也采用 567 题的滑动窗口思路，将每次能成功相等的窗口的起始位置都记录下来。

3、代码

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        if(s == null || s.length() == 0 || p == null || p.length() == 0){
            return new ArrayList<>();
        }

        Map<Character, Integer> window = new HashMap<>();
        Map<Character, Integer> need = new HashMap<>();

        int m = s.length(), n = p.length();
        if(m < n){
            return new ArrayList<>();
        }

        for (char c : p.toCharArray()) {
            need.put(c, need.getOrDefault(c, 0) + 1);
        }

        for(int i = 0; i < n; i++){
            window.put(s.charAt(i), window.getOrDefault(s.charAt(i), 0) + 1);
        }

        List<Integer> list = new ArrayList<>();
        if(window.equals(need)){
            list.add(0);
        }

        for(int i = n; i < m; i++){
            char removeChar = s.charAt(i - n);
            char addChar = s.charAt(i);
            window.put(removeChar, window.get(removeChar) - 1);
            if(window.get(removeChar) == 0){
                window.remove(removeChar);
            }
            window.put(addChar, window.getOrDefault(addChar, 0) + 1);

            if(window.equals(need)){
                list.add(i - n + 1);
            }
        }
        

        return list;
    }
}