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;
}
};