[LeetCode] 438. Find All Anagrams in a String

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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