【leetcode】1561. Maximum Number of Coins You Can Get

题目如下：

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
Constraints:

3 <= piles.length <= 10^5

piles.length % 3 == 0

1 <= piles[i] <= 10^4

解题思路：每次从piles中选出最大值, 次大值和最小值。最大值给Alice，最小值给Bob，自己拿次大值，直到piles中全部元素分配完成即可。

代码如下：

class Solution(object):
    def maxCoins(self, piles):
        """
        :type piles: List[int]
        :rtype: int
        """
        piles.sort(reverse=True)
        res = 0
        for i in range(1,len(piles) - len(piles)/3,2):
            res += piles[i]
        return res