本文主要是介绍[LeetCode] 438. Find All Anagrams in a String，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题：https://leetcode.com/problems/find-all-anagrams-in-a-string/description/
题目
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".思路
题目分析
题目要求 从 s 字符串中 找出 特殊子段的 起始位置。
特殊子段：该子段字母出现的 次数 与 p 字符串 出现的次数完全相同。
解题思路
1.生成 p 对应的窗口格式
2.对 s 生成一个 滑动窗口，滑动窗口用 dict 表示。每向前滑动一个窗口 ，右边字符 在dict中 value 减一，若value == 0 ，同时删除 key值。
3.对比 p 与 s 的窗口。
code
from collections import defaultdictclass Solution:def findAnagrams(self, s, p):""":type s: str:type p: str:rtype: List[int]"""res = []windict = defaultdict(lambda: 0)pdict = defaultdict(lambda: 0)plen = len(p)if plen >len(s):return resfor i in range(plen):pdict[p[i]] += 1for i in range(plen):windict[s[i]] += 1if windict.items() == pdict.items():res.append(0)for i in range(plen, len(s)):windict[s[i - plen]] -= 1if windict[s[i - plen]] == 0:del windict[s[i - plen]]windict[s[i]] += 1if windict.items() == pdict.items():res.append(i - plen + 1)return res这篇关于[LeetCode] 438. Find All Anagrams in a String的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
