LeetCode 914. Card Packet C ++ description

LeetCode 914. Card Packet C ++ description

Difficulty simply
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

class Solution {
public:
    bool hasGroupsSizeX(vector<int>& deck) {
        
        sort(deck.begin(),deck.end());
        for(int i = 2; i <= deck.size(); i++){
            //如果不能整除，则直接下一个
            if(deck.size() % i != 0)
                continue;
            bool flag = true;
            for(int j = 0; j < deck.size()/i; j++){
                for(int k = j * i; k < (j+1) * i && k < deck.size(); k++){
                    if(deck[k] != deck[j*i]){
                        //如果不和第一个数一样，则退出
                        flag = false;
                        break;
                    }
                    if(!flag)
                        break;
                }
            }
            if(flag)
                    return true;
        }
        return false;

    }
};