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: 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
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
