"Hash Table" 954. Two-fold pair array <LeetCode>

954. Double pairs array

Medium difficulty 32 Favorites sharing Switch to English to receive dynamic feedback

Given an integer array with an even length  , return Aonly if the  A "for each 0 <= i < len(A) / 2has  A[2 * i + 1] = 2 * A[2 * i]" is  satisfied after reorganization  true; otherwise, return  false.

Problem solution: classification + map (hash_map should be used, but I don't know why hash_map reports an error in LeetCode)

class Solution {
public:
    //只找绝对值比自己小的，8 4 2 16  会true
    bool canReorderDoubled(vector<int>& A) {
        int n=A.size();
        if(!n)
            return true;

        sort(A.begin(),A.end());
        map<int,int>mp;
        for(int i : A) mp[i]++; 
        
        //map有序 
        for(int i=0;i<n;i++)
        {
            int match=-1;
            if(mp[A[i]]>0)
            {
                if(A[i]>0)//正数 从小到大 找*2的
                    match=A[i]*2;
                else//负数和0  绝对值从大到小，找/2的
                {
                    if(A[i]&1)
                        return false;
                    match=A[i]/2;
                }
                if(mp[A[i]]<=mp[match])
                {
                    mp[match]-=mp[A[i]];
                    mp[A[i]]=0;
                }else
                    return false;
            }
        }
        return true;
    }
};

The application of the hash table is actually C++STL: hash_map most of the time, just learn to use it. If you want to understand the hash_map, you can visit: hash_map .