leetcode 438. Find All Anagrams in a String 一个简单的移动窗口问题

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”.

题意很简单，就是一个简单的移动窗口的问题，直接遍历比较窗口即可

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <cmath>using namespace std;class Solution {public:    vector<int> findAnagrams(string s, string p)     {        vector<int> res;        vector<int> win(26, 0) , tar(26, 0);        for (int i = 0; i < p.length(); i++)        {            win[s[i] - 'a']++;            tar[p[i] - 'a']++;        }        if (win == tar)            res.push_back(0);        for (int i = p.length(); i < s.length(); i++)        {            win[s[i] - 'a']++;            win[s[i-p.length()] - 'a']--;            if (win == tar)                res.push_back(i - p.length() + 1);        }        return res;    }};