LeetCode | 0914. X of a Kind in a Deck of Cards卡牌分组【Python】

深碍√TFBOYSˉ_ 2023-05-28 05:11 8阅读 0赞

> LeetCode 0914. X of a Kind in a Deck of Cards卡牌分组【Easy】【Python】【数学】

### Problem ###

[LeetCode][]

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: deck = [1,2,3,4,4,3,2,1]
    Output: true
    Explanation: Possible partition [1,1],[2,2],[3,3],[4,4].

**Example 2:**

Input: deck = [1,1,1,2,2,2,3,3]
    Output: false´
    Explanation: No possible partition.

**Example 3:**

Input: deck = [1]
    Output: false
    Explanation: No possible partition.

**Example 4:**

Input: deck = [1,1]
    Output: true
    Explanation: Possible partition [1,1].

**Example 5:**

Input: deck = [1,1,2,2,2,2]
    Output: true
    Explanation: Possible partition [1,1],[2,2],[2,2].

**Constraints:**

*  `1 <= deck.length <= 10^4`
 *  `0 <= deck[i] < 10^4`

### 问题 ###

[力扣][Link 1]

给定一副牌，每张牌上都写着一个整数。

此时，你需要选定一个数字 X，使我们可以将整副牌按下述规则分成 1 组或更多组：

*  每组都有 X 张牌。
 *  组内所有的牌上都写着相同的整数。

仅当你可选的 X >= 2 时返回 true。

**示例 1：**

输入：[1,2,3,4,4,3,2,1]
    输出：true
    解释：可行的分组是 [1,1]，[2,2]，[3,3]，[4,4]

**示例 2：**

输入：[1,1,1,2,2,2,3,3]
    输出：false
    解释：没有满足要求的分组。

**示例 3：**

输入：[1]
    输出：false
    解释：没有满足要求的分组。

**示例 4：**

输入：[1,1]
    输出：true
    解释：可行的分组是 [1,1]

**示例 5：**

输入：[1,1,2,2,2,2]
    输出：true
    解释：可行的分组是 [1,1]，[2,2]，[2,2]

**提示：**

*  `1 <= deck.length <= 10^4`
 *  `0 <= deck[i] < 10^4`

### 思路 ###

**数学**

直接调用 Python 的 collections 库，Counter 统计。
    然后计算所有整数出现次数的最大公约数。

##### Python3代码 #####

from typing import List
    import collections
    import math
    
    class Solution:
        def hasGroupsSizeX(self, deck: List[int]) -> bool:
            # Counter统计出来是一个字典
            c = list(collections.Counter(deck).values())
            res = c[0]
            # 求所有数的最大公约数
            for x in c:
                res = math.gcd(res, x)
            return res > 1

### GitHub链接 ###

[Python][]

[LeetCode]: https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/
[Link 1]: https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/
[Python]: https://github.com/Wonz5130/LeetCode-Solutions/blob/master/solutions/0914-X-of-a-Kind-in-a-Deck-of-Cards/0914.py