Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

 

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".

 

Constraints:

• 1 <= s.length, p.length <= 3 * 104
• s and p consist of lowercase English letters.

利用hash表进行比较，只要包含的相同的子序列字符即可，然后利用滑动窗口进行比较。

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        // 如何拆解string p 构成需要匹配的anagrams
        // 变位词只要保证存在这个词即可 不需要顺序 利用hash表进行对比
        vector<int> vecRes={};
        int nLs=s.size();
        int nLp=p.size();
        if(nLs<nLp) return vecRes;
        // 两个hash表 进行存储
        vector<int> vecPoint(26,0),vecSlide(26,0);
        for(int i=0;i<nLp;++i){
            vecPoint[p[i]-'a']+=1;
            vecSlide[s[i]-'a']+=1;
        }
        if(vecPoint==vecSlide){
            vecRes.push_back(0);
        }
        int index=0;
        for(index;index<nLs-nLp;index++){
            //cout<<index<<endl; 
            // move this slide window
            vecSlide[s[index]-'a']-=1;
            vecSlide[s[index+nLp]-'a']+=1;
            if(vecPoint==vecSlide){
                vecRes.push_back(index+1);
            }
        }
        return vecRes;
        
    }
};