[leetcode: Python]438. Find All Anagrams in a String

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".

找到一个字符串里所有的字谜
方法一：295ms

from collections import Counterclass Solution(object):    def findAnagrams(self, s, p):        """        :type s: str        :type p: str        :rtype: List[int]        """        ans = []        ls, lp = len(s), len(p)        count  = lp        cp = collections.Counter(p)        for i in range(ls):            if cp[s[i]] >= 1:                count -= 1            cp[s[i]] -= 1            if i >= lp:                if cp[s[i-lp]] >= 0:                    count += 1                cp[s[i-lp]] += 1            if count == 0:                ans.append(i - lp + 1)        return ans

方法二：146ms

from collections import defaultdictclass Solution(object):    def findAnagrams(self, s, p):        """        :type s: str        :type p: str        :rtype: List[int]        """        res = []        if not p or not s:            return res        lenP = len(p)        lenS = len(s)        if lenS < lenP:            return res        hP = defaultdict(int)        for ch in p:            hP[ch] += 1        count = len(hP.keys())        for i in xrange(lenP):            ch = s[i]            if ch in hP:                hP[ch] -= 1                if hP[ch] == 0:                    count -= 1        if count == 0:            res.append(0)        left = 0        right = lenP        while right < lenS:            ch = s[right]            if ch in hP:                hP[ch] -= 1                if hP[ch] == 0:                    count -= 1            ch = s[left]            if ch in hP:                hP[ch] += 1                if hP[ch] == 1:                    count += 1            left += 1            right += 1            if count == 0:                res.append(left)        return res