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
wherepiles[i]
is the number of coins in theith
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: 4Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4] Output: 18Constraints:
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