一、题目

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

 

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".
 

Constraints:

1 <= s.length, p.length <= 3 * 104
s and p consist of lowercase English letters.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-all-anagrams-in-a-string
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二、思路

套模板，debug的地方就四个或者五个，弄明白每个关键点到底应该怎么写就ok了。

二、自己写的

1.代码

第一次写

class Solution {
    public List<Integer> findAnagrams(String s, String p) {

        List<Integer> res = new ArrayList<Integer>();

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

        int left = 0, right = 0;
        int valid = 0;
        int len = Integer.MAX_VALUE;

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

        //第一个debug点，考虑什么时候窗口右移
        while (right < s.length()) {

            char c = s.charAt(right);
            right++;
            len = right - left;

            //第二个debug点，什么时候window和valid“变大”
            if (need.containsKey(c)) {
                window.put(c, window.getOrDefault(c, 0) + 1);
                if (need.get(c).equals(window.get(c)))
                    valid++;
            }

            //第三个debug点，什么时候开始窗口左移
            while (len >= p.length()) {
                //什么情况下满足要求
                if ((len == p.length()) && (valid == need.size()))
                    res.add(left);

                char d = s.charAt(left);
                left++;
                len = right - left;
                
                //第四个debug点，什么时候window和valid“变小”
                if (need.containsKey(d)) {
                    if (need.get(d).equals(window.get(d)))
                        valid--;
                    window.put(d, window.get(d) - 1);
                }
            }
        }
        return res;
    }
}

//正确，弄清楚那几个点的逻辑就ok。

三、标准答案

1.方法一（滑动窗口）

class Solution {
    public List<Integer> findAnagrams(String s, String p) {

        List<Integer> res = new ArrayList<Integer>();

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

        int left = 0, right = 0;
        int valid = 0;
        int len = Integer.MAX_VALUE;

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

        //第一个debug点，考虑什么时候窗口右移
        while (right < s.length()) {

            char c = s.charAt(right);
            right++;
            len = right - left;

            //第二个debug点，什么时候window和valid“变大”
            if (need.containsKey(c)) {
                window.put(c, window.getOrDefault(c, 0) + 1);
                if (need.get(c).equals(window.get(c)))
                    valid++;
            }

            //第三个debug点，什么时候开始窗口左移
            while (len >= p.length()) {
                //什么情况下满足要求
                if ((len == p.length()) && (valid == need.size()))
                    res.add(left);

                char d = s.charAt(left);
                left++;
                len = right - left;
                
                //第四个debug点，什么时候window和valid“变小”
                if (need.containsKey(d)) {
                    if (need.get(d).equals(window.get(d)))
                        valid--;
                    window.put(d, window.get(d) - 1);
                }
            }
        }
        return res;
    }
}