*[LeetCode] 438. Find All Anagrams in a String_find all anagrams in a string javascript-优快云博客

博客围绕LeetCode上查找字符串变位词起始索引的题目展开。指出若每次选中字符串排序对比效率低，可采用滑动窗口方法比对，利用字典处理无序情况，同时要约束窗口长度，还提及了程序代码但未给出具体内容。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目内容

https://leetcode-cn.com/problems/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 pwill 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".

题目思路

本题如果采用每次都进行选中字符串排序对比的话，效率会非常的低，所以我们可以考虑使用滑动窗口的方法进行比对。因为是无序的，所以可以采用字典的方法。同时要保证对窗口的长度进行约束。

程序代码

class Solution(object):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        dp,ds={},{}
        for i in p:
            dp[i]=dp.get(i,0)+1
        res=[]
        lp=len(p)
        for i,val in enumerate(s):
            ds[val]=ds.get(val,0)+1
            if ds==dp:
                res.append(i-lp+1)#回退到本次开头
            if i-lp+1>=0:
                ds[s[i-lp+1]]=ds.get(s[i-lp+1],0)-1
                if ds[s[i-lp+1]]==0:
                    del ds[s[i-lp+1]]
        return res