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  • 1561. Maximum Number of Coins You Can Get

    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。

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/13584952.html
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