[leetcode] 914. X of a Kind in a Deck of Cards

[leetcode] 914. X of a Kind in a Deck of Cards

Description

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: trueExplanation: 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: falseExplanation: No possible partition.

Example 4:

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

Example 5:

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

Constraints:

1 <= deck.length <= 10^40 <= deck[i] < 10^4

分析

题目的意思是：把一个数组里面的数字平均分成k个组，每个组里面的数字是一样的，这在leetcode里面是一道easy题目，我也没做出来，太尴尬了，答案是先统计每个数的频率，然后遍历寻找合适的X，如果找到就能够分成功，找不到，就是False了。

代码

class Solution: def hasGroupsSizeX(self, deck: List[int]) -> bool: count=collections.Counter(deck) N=len(deck) for X in range(2,N+1): if(N%X==0): flag=True for v in count.values(): if(v%X!=0): flag=False if(flag): return True return False

参考文献

​​[LeetCode] Approach 1: Brute Force​​

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