描述
There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
- In each step, you will choose any 3 piles of coins (not necessarily consecutive).
- Of your choice, Alice will pick the pile with the maximum number of coins.
- You will pick the next pile with maximum number of coins.
- Your friend Bob will pick the last pile.
- Repeat until there are no more piles of coins.
Given an array of integers piles where piles[i] is the number of coins in the ith pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Note:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
解析
根据题意，其实游戏步骤就是三步，我先随便挑三种硬币类型，Alice 取最多的那种，我取次多的那种，Bob 取最少的那种，重复这个过程，问每次我可能得到的最多的硬币数量是多少。其实规则很简单，通过找规律就会发现，假如 piles 是 [1,2,4,2,7,8]，则步骤如下取结果最大：
[1,7,8] 取 7
[2,2,4] 取 2
假如 piles 是 [1,2,3,4,5,6,7,8,9] ，
[1,8,9] 取 8
[2,6,7] 取 6
[3,4,5] 取 4
会使得最后的结果最大，可以总结出规律，一共我可以取 len(piles)//3 次，每次取 piles 升序排列结果的倒数第二个元素、第四个元素、第六个元素。。。的值，会使得最后的结果最大。
解答
class Solution(object):
def maxCoins(self, piles):
"""
:type piles: List[int]
:rtype: int
"""
piles.sort()
res = 0
n = len(piles)
for i in range(len(piles)//3):
res += piles[n-(i+1)*2]
return res
运行结果
Runtime: 544 ms, faster than 74.36% of Python online submissions for Maximum Number of Coins You Can Get.
Memory Usage: 23.3 MB, less than 68.80% of Python online submissions for Maximum Number of Coins You Can Get.
原题链接：leetcode.com/problems/ma…
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