438. Find All Anagrams in a String

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".


如果s的子串包含p串（不限顺序），则输出子串的起始索引

C++（43ms）：

 1 class Solution {
 2 public:
 3     vector<int> findAnagrams(string s, string p) {
 4         vector<int> res ;
 5         vector<int> hash(256) ;
 6         for(auto c : p){
 7             hash[c]++ ;
 8         }
 9         int left = 0 ;
10         int right = 0 ;
11         int count = p.size() ;
12         while(right < s.size()){
13             if (hash[s[right++]]-- >= 1){
14                 count-- ;
15             }
16             if (count == 0){
17                 res.push_back(left) ;
18             }
19             if (right - left == p.size() && ++hash[s[left++]] >= 1){
20                 count++ ;
21             }
22         }
23         return res ;
24     }
25 };