leetcode 914. card packet (the most common factors play gcd)

Given a deck of cards, each card are written on an integer.

At this point, you need to select a digital X, so that we can be full decks of cards according to the following rules into one or more groups:

Each group has X cards.
Within the group are written the same integer on all the cards.
> = 2 returns true only if you optional X.

 

Example 1:

Input: [1,2,3,4,4,3,2,1]
Output: true
interpretation: grouping is possible [1,1], [2,2], [3,3], [4,4 ]
example 2:

Input: [1,1,1,2,2,2,3,3]
Output: false
interpretation: the packet does not satisfy the requirements.
Example 3:

Input: [1]
Output: false
interpretation: the packet does not satisfy the requirements.
Example 4:

Input: [1,1]
Output: true
interpretation: grouping is possible [1,1]
Example 5:

Input: [1,1,2,2,2,2]
Output: true
interpretation: grouping is possible [1,1], [2,2], [2,2]

prompt:

1 <= deck.length <= 10000
0 <= deck[i] < 10000

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards

class Solution {
public:
    int gcd(int a,int b){
        return b==0?a:gcd(b,a%b);
    }
    bool hasGroupsSizeX(vector<int>& deck) {
        int a[10010];
        memset(a,0,sizeof(a));
        int num;
        int maxx=0;
        for(int i=0;i<deck.size();i++){
            a[deck[i]]++;
            num=a[deck[i]];
            maxx=max(maxx,deck[i]);
        }
        for(int i=0;i<=maxx;i++){
            if(a[i]!=0)num=gcd(num,a[i]);
        }
        if(num==1)return false;
        return true;
    }
};