LeetCode 438. Find All Anagrams in a String--字符串-滑动窗口--C++,Python解法

题目地址：Find All Anagrams in a String - LeetCode

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".

这道题目的意思很容易懂，就是看s的字母是否是p字符串的排列组合。如果把p字符串的所有可能都列出来，太多了，于是我想到使用counter。

Python解法如下：

from collections import Counter
class Solution:def findAnagrams(self, s: str, p: str) -> List[int]:res = []p_set = dict(Counter(p))lens = len(s)lenp = len(p)if lens < lenp:return resfor i in range(0, lens-lenp+1):s_set = dict(Counter(s[i:i+lenp]))if s_set == p_set:res.append(i)return res

但上面的原始的Python解法会超时，因为有很多重复的工作，使用滑动窗口来减少工作量：

from collections import Counter
class Solution:def findAnagrams(self, s: str, p: str) -> List[int]:res = []p_set = Counter(p)lens = len(s)lenp = len(p)if lens < lenp:return ress_set = Counter(s[0:lenp-1])for i in range(lenp-1, lens):s_set[s[i]] += 1  # include a new char in the windowstart = i-lenp+1if s_set == p_set:res.append(start)s_set[s[start]] -= 1 # decrease the count of oldest char in the windowif s_set[s[start]] == 0:del s_set[s[start]]  # remove the count if it is 0return res

上面的解法可以通过了，但可以更快，因为中间还是做了很多无用功，如果能快速跳过不存在的字符，可以更快。
更快的解法：

from collections import Counterclass Solution:def findAnagrams(self, s: str, p: str) -> List[int]:res = []pkey = set(p)p_set = Counter(p)lens = len(s)lenp = len(p)if lens < lenp:return ress_set = Counter(s[0:lenp-1])i = lenp-1while i < lens:if s[i] not in pkey:start = i+1i = start+lenp-1if i >= lens:breaks_set = Counter(s[start:i])continues_set[s[i]] += 1  # include a new char in the windowstart = i-lenp+1if s_set == p_set:res.append(start)# decrease the count of oldest char in the windows_set[s[start]] -= 1if s_set[s[start]] == 0:del s_set[s[start]]  # remove the count if it is 0i += 1return res

C++解法如下：

class Solution {public:vector<int> findAnagrams(string s, string p) {vector<int> pv(26,0), sv(26,0), res;if(s.size() < p.size())return res;// fill pv, vector of counters for pattern string and sv, vector of counters for the sliding windowfor(int i = 0; i < p.size(); ++i){++pv[p[i]-'a'];++sv[s[i]-'a'];}if(pv == sv)res.push_back(0);//here window is moving from left to right across the string. //window size is p.size(), so s.size()-p.size() moves are made for(int i = p.size(); i < s.size(); ++i) {// window extends one step to the right. counter for s[i] is incremented ++sv[s[i]-'a'];// since we added one element to the right, // one element to the left should be discarded. //counter for s[i-p.size()] is decremented--sv[s[i-p.size()]-'a']; // if after move to the right the anagram can be composed, // add new position of window's left point to the result if(pv == sv)  // this comparison takes O(26), i.e O(1), since both vectors are of fixed size 26. Total complexity O(n)*O(1) = O(n)res.push_back(i-p.size()+1);}return res;}
};

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