Last Updated on 2023-11-24 by Clay
題目
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
3piles 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 the 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 that 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 <= 105piles.length % 3 == 01 <= piles[i] <= 104
解題思路
這個題目乍看之下似乎很複雜，但想清楚大家拿硬幣的順序是固定的，就可以用貪婪演算法直接計算我們所能拿到的硬幣最大數量。
由於我們一定是第二位拿的，所以可以想像成每一輪我們都固定拿第二大的硬幣堆，而最後選的一定要給他最小的硬幣堆 —— 所以我們所能拿的最大值，就是每一輪都組成 (當前最大硬幣堆, 當前第二大硬幣堆, 當前最小硬幣堆) 這樣的硬幣堆選取。
以 Example 1 為例：[2,4,1,2,7,8] 我們會先重排成 [1, 2, 2, 4, 7, 8]，接著第一輪我們會選擇 (1, 7, 8) 的組合，我們第二位挑選所以只能挑 7；而第二輪我們只能選剩下的 (2, 2, 4)，我們只能拿 2，所以答案是 7 + 2 = 9 —— 就是我們能拿到的最多硬幣數量組合。
C++ 範例程式碼
class Solution {
public:
int maxCoins(vector<int>& piles) {
// Init
int coins = 0;
int left = 0;
int right = piles.size() - 1;
// Sort
sort(piles.begin(), piles.end());
// Count
while (left < right - 1) {
coins += piles[right-1];
++left;
right -= 2;
}
return coins;
}
};Python 範例程式碼
class Solution:
def maxCoins(self, piles: List[int]) -> int:
# Init
coins = 0
left = 0
right = len(piles) - 1
# Sort
piles = sorted(piles)
# Count
while left < right - 1:
coins += piles[right - 1]
right -= 2
left += 1
return coins