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
class Solution { public int maxCoins(int[] piles) { Arrays.sort(piles); int result = 0; int n = piles.length; int l = 0, r = n - 1; int[] res = new int[n]; int ind = 0; while(l < r) { res[ind++] = piles[r--]; res[ind++] = piles[r--]; res[ind++] = piles[l++]; } for(int i = 1; i < n; i += 3) result += res[i]; return result; } }
Sort, 每次取两个最大的和一个最小的重新排列,最后每隔3个把中间那个取出来。
class Solution { public int maxCoins(int[] piles) { Arrays.sort(piles); int result = 0; int n = piles.length; int times = n / 3; int ind = n - 2; while(times-- > 0) { result += piles[ind]; ind -= 2; } return result; } }
啊?排完序后直接从倒数第二个开始每隔一个取,取n / 3次。