Find All Anagrams in a String

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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:

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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:

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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

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from collections import Counter
class Solution(object):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        res  = []
        p_len = len(p)
        s_len = len(s)
        p_counter = Counter(p)
        s_tmp_counter = Counter(s[:p_len])
        if s_tmp_counter == p_counter:
            res.append(0)
        for i in range(p_len,s_len):
            s_tmp_counter[s[i-p_len]] -= 1
            if s_tmp_counter[s[i-p_len]] == 0:
                del s_tmp_counter[s[i-p_len]]
            s_tmp_counter[s[i]] += 1
            if s_tmp_counter == p_counter:
                res.append(i+1-p_len)
        return res