問題描述

914. X of a Kind in a Deck of Cards

In a deck of cards, each card has an integer written on it.

Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where:

• Each group has exactly X cards.
• All the cards in each group have the same integer.

Example 1:

Input: [1,2,3,4,4,3,2,1]

Output: true

Explanation: Possible partition [1,1],[2,2],[3,3],[4,4]

Example 2:

Input: [1,1,1,2,2,2,3,3]

Output: false

Explanation: No possible partition.

Example 3:

Input: [1]

Output: false

Explanation: No possible partition.

Example 4:

Input: [1,1]
 

Output: true

Explanation: Possible partition [1,1]

Example 5:

Input: [1,1,2,2,2,2]

Output: true

Explanation: Possible partition [1,1],[2,2],[2,2]

Note:

1. 1 <= deck.length <= 10000
2. 0 <= deck[i] < 10000

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題意

給定一個數列A，問數列A能不能分成n個部分（n>=1），每個部分元素個數>=2,且每個元素相同

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思路

用map記錄數列A中每個元素出現的次數，問題轉化為求這些次數的公因子。如果有大於等於2的公因子，則返回true。求公因子時採用列舉從2到最小數依次判斷整除性的辦法。

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程式碼

class Solution {
public:
    bool hasGroupsSizeX(vector<int>& deck) {
        map<int, int> mp;
        map<int, int>::iterator it;
        int i, len = deck.size(), xmin = 0x3f3f3f3f, x;
        bool is_true = true;
        for (i=0; i<len; i++)
        {
            mp[deck[i]]++;
        }
        for (it=mp.begin(); it!=mp.end(); it++)
        {
            xmin = min(it->second, xmin);
        }
        for (x = 2; x <= xmin; x++)
        {
            is_true = true;
            for (it=mp.begin(); it!=mp.end(); it++)
            {
                if ((it->second)%x != 0)
                {
                    is_true = false;
                }
            }
            if (is_true)
            {
                return true;
            }
        }
        return false;
    }
};