【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