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
你可以获得的最大硬币数目。
有 3n 堆数目不一的硬币，你和你的朋友们打算按以下方式分硬币：
每一轮中，你将会选出 任意 3 堆硬币（不一定连续）。
Alice 将会取走硬币数量最多的那一堆。
你将会取走硬币数量第二多的那一堆。
Bob 将会取走最后一堆。
重复这个过程，直到没有更多硬币。
给你一个整数数组 piles ，其中 piles[i] 是第 i 堆中硬币的数目。返回你可以获得的最大硬币数目。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-number-of-coins-you-can-get
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
思路是排序 + two pointer。这道题我们必须排序，否则是拿不到次多的那一堆的。排序完之后，我们用左右两个指针逼近中间，左指针 i 指向的是每次选取的三堆硬币中最少的，右指针 j 指向的是每次选取的三堆硬币中最多的，j - 1 指向的是次多的。要选取的次数是piles.length / 3。
时间O(nlogn)
空间O(1)
Java实现
1 class Solution { 2 public int maxCoins(int[] piles) { 3 Arrays.sort(piles); 4 int len = piles.length; 5 int count = len / 3; 6 int res = 0; 7 int i = 0; 8 int j = len - 1; 9 while (i < j && count != 0) { 10 if (j - 1 >= i) { 11 res += piles[j - 1]; 12 count--; 13 } 14 i++; 15 j -= 2; 16 } 17 return res; 18 } 19 }