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）：

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> res ;
        vector<int> hash(256) ;
        for(auto c : p){
            hash[c]++ ;
        }
        int left = 0 ;
        int right = 0 ;
        int count = p.size() ;
        while(right < s.size()){
            if (hash[s[right++]]-- >= 1){
                count-- ;
            }
            if (count == 0){
                res.push_back(left) ;
            }
            if (right - left == p.size() && ++hash[s[left++]] >= 1){
                count++ ;
            }
        }
        return res ;
    }
};