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

客官°小女子只卖身不卖艺 2022-06-04 09:17 120阅读 0赞

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;
        }
    };