遍历+哈希表
O(Ns + Np)

class Solution {
public:
    bool compare(vector<int> v1, vector<int> v2){
        for (int i=0;i<26;i++) {
            if (v1[i]!=v2[i]) return false;
        }
        return true;
    }
    vector<int> findAnagrams(string s, string p) {
        if (s.size()<p.size()) return {};
        vector<int> ans;
        vector<int> hash_p(26,0), hash_s(26,0);
        // unordered_set<char> set_p, set_s;
        for (auto c:p) hash_p[c-'a']++;
        int len_p = p.size();

        for (int i =0;i<len_p;i++) {
            char j = s[i] - 'a';
            hash_s[j]++;
        }
        if (compare(hash_p, hash_s)) ans.push_back(0);
        int l=0, r = len_p-1;
        while (r+1<s.size()){
            r++;
            int j1 = s[l]-'a', j2 = s[r]-'a';
            hash_s[j1]--;
            hash_s[j2]++;
            l++;
            if (compare(hash_s,hash_p)) ans.push_back(l);
        }
        return ans;
    }
};