标签：counter turn tput append http anagrams let port mos
https://leetcode.com/problems/find-all-anagrams-in-a-string/#/description
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".
Sol:
class Solution(object): def findAnagrams(self, s, p): """ :type s: str :type p: str :rtype: List[int] """ from collections import Counter res = [] pCounter = Counter(p) sCounter = Counter(s[:len(p)-1]) for i in range(len(p)-1,len(s)): sCounter[s[i]] += 1 # include a new char in the window if sCounter == pCounter: # This step is O(1), since there are at most 26 English letters res.append(i-len(p)+1) # append the starting index sCounter[s[i-len(p)+1]] -= 1 # decrease the count of oldest char in the window if sCounter[s[i-len(p)+1]] == 0: del sCounter[s[i-len(p)+1]] # remove the count if it is 0 return res
Note:
1 counter usage:
print collections.Counter([‘a‘, ‘b‘, ‘c‘, ‘a‘, ‘b‘, ‘b‘])
438. Find All Anagrams in a String
标签：counter turn tput append http anagrams let port mos
原文地址：http://www.cnblogs.com/prmlab/p/6978073.html
