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‎README.adoc

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//|Easy
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//|{counter:codes}
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//|{leetcode_base_url}/maximum-number-of-coins-you-can-get/[1561. Maximum Number of Coins You Can Get^]
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//|{source_base_url}/_1561_MaximumNumberOfCoinsYouCanGet.java[Java]
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//|{doc_base_url}/1561-maximum-number-of-coins-you-can-get.adoc[题解]
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|{leetcode_base_url}/maximum-number-of-coins-you-can-get/[1561. Maximum Number of Coins You Can Get^]
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|{source_base_url}/_1561_MaximumNumberOfCoinsYouCanGet.java[Java]
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|{doc_base_url}/1561-maximum-number-of-coins-you-can-get.adoc[题解]
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//|{leetcode_base_url}/find-latest-group-of-size-m/[1562. Find Latest Group of Size M^]

‎docs/1561-maximum-number-of-coins-you-can-get.adoc

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[#1561-maximum-number-of-coins-you-can-get]
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= 1561. Maximum Number of Coins You Can Get
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= 1561. 你可以获得的最大硬币数目
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{leetcode}/problems/maximum-number-of-coins-you-can-get/[LeetCode - 1561. Maximum Number of Coins You Can Get ^]
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https://leetcode.cn/problems/maximum-number-of-coins-you-can-get/[LeetCode - 1561. 你可以获得的最大硬币数目 ^]
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There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
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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：
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* 每一轮中，你将会选出 *任意* 3 堆硬币（不一定连续）。
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* Alice 将会取走硬币数量最多的那一堆。
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* 你将会取走硬币数量第二多的那一堆。
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* Bob 将会取走最后一堆。
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* 重复这个过程，直到没有更多硬币。
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* In each step, you will choose *any *`3` piles of coins (not necessarily consecutive).
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* Of your choice, Alice will pick the pile with the maximum number of coins.
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* You will pick the next pile with the maximum number of coins.
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* Your friend Bob will pick the last pile.
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* Repeat until there are no more piles of coins.
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给你一个整数数组 `piles` ，其中 `piles[i]` 是第 `i` 堆中硬币的数目。
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返回你可以获得的最大硬币数目。
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Given an array of integers `piles` where `piles[i]` is the number of coins in the `i^th^` pile.
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Return the maximum number of coins that you can have.
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*示例 1：*
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*Example 1:*
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....
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输入：piles = [2,4,1,2,7,8]
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输出：9
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解释：选出 (2, 7, 8) ，Alice 取走 8 枚硬币的那堆，你取走 7 枚硬币的那堆，Bob 取走最后一堆。
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选出 (1, 2, 4) , Alice 取走 4 枚硬币的那堆，你取走 2 枚硬币的那堆，Bob 取走最后一堆。
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你可以获得的最大硬币数目：7 + 2 = 9.
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考虑另外一种情况，如果选出的是 (1, 2, 8) 和 (2, 4, 7) ，你就只能得到 2 + 4 = 6 枚硬币，这不是最优解。
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....
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[subs="verbatim,quotes"]
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----
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*Input:* piles = [2,4,1,2,7,8]
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*Output:* 9
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*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.
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Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with *2* coins and Bob the last one.
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The maximum number of coins which you can have are: 7 + 2 = 9.
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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.
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----
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*示例 2：*
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*Example 2:*
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....
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输入：piles = [2,4,5]
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输出：4
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....
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[subs="verbatim,quotes"]
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----
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*Input:* piles = [2,4,5]
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*Output:* 4
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----
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*示例 3：*
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*Example 3:*
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....
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输入：piles = [9,8,7,6,5,1,2,3,4]
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输出：18
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....
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[subs="verbatim,quotes"]
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----
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*Input:* piles = [9,8,7,6,5,1,2,3,4]
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*Output:* 18
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----
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*Constraints:*
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*提示：*
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* `3 <= piles.length <= 10^5^`
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* `3 \<= piles.length <= 10^5^`
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* `piles.length % 3 == 0`
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* `1 <= piles[i] <= 10^4^`
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* `1 \<= piles[i] \<= 10^4^`
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== 思路分析
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因为每次最高都被 Alice 拿走，想要获取最多硬币，那就要取一个尽可能靠近最大值的数量，对数组排序，从后面取两个，从前面取一个，这样中间的值之和就可以取到最大的值。
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TIP: 把下标写出来，更容易找出规律。可以让条件写的更简单。
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[[src-1561]]
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== 参考资料
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. https://leetcode.cn/problems/maximum-number-of-coins-you-can-get/solutions/409109/ni-ke-yi-huo-de-de-zui-da-ying-bi-shu-mu-by-leetco/[1561. 你可以获得的最大硬币数目 - 官方题解^]
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. https://leetcode.cn/problems/maximum-number-of-coins-you-can-get/solutions/3035907/tan-xin-jian-ji-xie-fa-pythonjavaccgojsr-2ptn/[1561. 你可以获得的最大硬币数目 - 贪心，简洁写法^]

‎docs/index.adoc

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‎logbook/202503.adoc

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|{doc_base_url}/1410-html-entity-parser.adoc[题解]
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|✅ 哈希 + 字符串
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|{counter:codes2503}
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|{leetcode_base_url}/maximum-number-of-coins-you-can-get/[1561. 你可以获得的最大硬币数目^]
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|{doc_base_url}/1561-maximum-number-of-coins-you-can-get.adoc[题解]
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|✅ 贪心。每次去第二大的数字即可。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_1561_MaximumNumberOfCoinsYouCanGet.java

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package com.diguage.algo.leetcode;
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import java.util.Arrays;
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public class _1561_MaximumNumberOfCoinsYouCanGet {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-05-22 23:28:11
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*/
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public int maxCoins(int[] piles) {
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Arrays.sort(piles);
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int result = 0;
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// TODO 有更简单的写法，尝试把坐标写出来，去寻找规律。
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for (int i = piles.length - 2, cnt = piles.length / 3;
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cnt > 0 && i > 0; i -= 2, cnt--) {
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result += piles[i];
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}
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return result;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _1561_MaximumNumberOfCoinsYouCanGet()
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.maxCoins(new int[]{2, 4, 1, 2, 7, 8});
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}
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}

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