【LeetCode】438. Find All Anagrams in a String 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/find-all-anagrams-in-a-string/description/

题目描述

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".

题目大意

在s中有多少个位置，从这个位置开始和p等长度的子字符串中，所包含的字符和p是一样的。

解题方法

滑动窗口

这个题考的是时间复杂度。如果判断两个切片是否是排列组合的话，时间复杂度略高，会超时。

能AC的做法是用了一个滑动窗口，每次进入窗口的字符的个数+1，超出滑动窗口的字符个数-1.

这样就一遍就搞定了，而且不用每个切片都算是不是一个排列组合。

Counter大法好，判断两个字符串是否是排列组合直接统计词频然后==判断即可。

注意如果一个词出现的次数是0，那么需要从Counter中移除，因为Counter({‘a’: 0, ‘b’: 1}) w不等于Counter({‘b’: 1})。

from collections import Counter

class Solution(object):

    def findAnagrams(self, s, p):

        """

        :type s: str

        :type p: str

        :rtype: List[int]

        """

        answer = []

        m, n = len(s), len(p)

        if m < n:

            return answer

        pCounter = Counter(p)

        sCounter = Counter(s[:n-1])

        index = 0

        for index in xrange(n - 1, m):

                sCounter[s[index]] += 1

                if sCounter == pCounter:

                    answer.append(index - n + 1)

                sCounter[s[index - n + 1]] -= 1

                if sCounter[s[index - n + 1]] == 0:

                    del sCounter[s[index - n + 1]]

        return answer

双指针

二刷的时候也是滑动窗口，但是使用的是双指针的解法，看上去好像没有上面这个方法更简单。

class Solution(object):

    def findAnagrams(self, s, p):

        """

        :type s: str

        :type p: str

        :rtype: List[int]

        """

        count = collections.Counter()

        M, N = len(s), len(p)

        left, right = 0, 0

        pcount = collections.Counter(p)

        res = []

        while right < M:

            count[s[right]] += 1

            if right - left + 1 == N:

                if count == pcount:

                    res.append(left)

                count[s[left]] -= 1

                if count[s[left]] == 0:

                    del count[s[left]]

                left += 1

            right += 1

        return res

日期

2018 年 1 月 27 日
2018 年 11 月 24 日 —— 周六快乐