LeetCode #438: Find All Anagrams in a String

最新推荐文章于 2025-05-18 16:01:12 发布

junchen1992

最新推荐文章于 2025-05-18 16:01:12 发布

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本文介绍了一种利用滑动窗口算法解决字符串中寻找所有目标子串字母异位词起始索引的问题。通过具体例子展示了算法的实现过程，并提供了一段Python代码作为示例。

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Problem Statement

(Source) 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".

Solution

A “Sliding Window” like algorithm.

class Solution(object):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        from collections import Counter

        res = []
        n1, n2 = len(s), len(p)
        cnt1, cnt2 = Counter(s[:n2]), Counter(p)
        for index in xrange(n1 - n2 + 1):
            if cnt1 == cnt2:
                res.append(index)
            if index == n1 - n2:
                break
            cnt1[s[index]] -= 1
            cnt1[s[index + n2]] += 1
            if cnt1[s[index]] == 0:
                cnt1.pop(s[index])
        return res

[To be optimised]