[LeetCode] 1753. Maximum Score From Removing Stones
You are playing a solitaire game with three piles of stones of sizes a
, b
, and c
respectively. Each turn you choose two different non-empty piles, take one stone from each, and add 1
point to your score. The game stops when there are fewer than two non-empty piles (meaning there are no more available moves).
Given three integers a
, b
, and c
, return the maximum score you can get.
Example 1:
Input: a = 2, b = 4, c = 6 Output: 6 Explanation: The starting state is (2, 4, 6). One optimal set of moves is: - Take from 1st and 3rd piles, state is now (1, 4, 5) - Take from 1st and 3rd piles, state is now (0, 4, 4) - Take from 2nd and 3rd piles, state is now (0, 3, 3) - Take from 2nd and 3rd piles, state is now (0, 2, 2) - Take from 2nd and 3rd piles, state is now (0, 1, 1) - Take from 2nd and 3rd piles, state is now (0, 0, 0) There are fewer than two non-empty piles, so the game ends. Total: 6 points.
Example 2:
Input: a = 4, b = 4, c = 6 Output: 7 Explanation: The starting state is (4, 4, 6). One optimal set of moves is: - Take from 1st and 2nd piles, state is now (3, 3, 6) - Take from 1st and 3rd piles, state is now (2, 3, 5) - Take from 1st and 3rd piles, state is now (1, 3, 4) - Take from 1st and 3rd piles, state is now (0, 3, 3) - Take from 2nd and 3rd piles, state is now (0, 2, 2) - Take from 2nd and 3rd piles, state is now (0, 1, 1) - Take from 2nd and 3rd piles, state is now (0, 0, 0) There are fewer than two non-empty piles, so the game ends. Total: 7 points.
Example 3:
Input: a = 1, b = 8, c = 8 Output: 8 Explanation: One optimal set of moves is to take from the 2nd and 3rd piles for 8 turns until they are empty. After that, there are fewer than two non-empty piles, so the game ends.
Constraints:
1 <= a, b, c <= 105
移除石子的最大得分。
你正在玩一个单人游戏,面前放置着大小分别为 a、b 和 c 的 三堆 石子。
每回合你都要从两个 不同的非空堆 中取出一颗石子,并在得分上加 1 分。当存在 两个或更多 的空堆时,游戏停止。
给你三个整数 a 、b 和 c ,返回可以得到的 最大分数 。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/maximum-score-from-removing-stones
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这道题的思路很容易想到,是贪心,但是做法有好几种。比较容易想到的做法是用 priority queue,但是代码运行起来很慢,这里我提供一个递归的代码,算是次优解吧。
根据题目的规则,每次需要从两个不同的堆中各拿出一个石子。为了让得分尽可能大,我这里采取的方案是优先从两个数量多的堆中取石子,所以我会先对 a,b,c 排序,确保 a > b > c。这样我递归的时候,就优先从 a 和 b 中取石子。
时间 - 应该是石子数量第二多的那一堆的数量,因为只要第二多的那一堆取完了,操作就停止了
空间O(n) - 递归栈空间
Java实现
1 class Solution { 2 public int maximumScore(int a, int b, int c) { 3 // make sure a > b > c 4 if (a < b) { 5 return maximumScore(b, a, c); 6 } 7 if (b < c) { 8 return maximumScore(a, c, b); 9 } 10 return b == 0 ? 0 : 1 + maximumScore(a - 1, b - 1, c); 11 } 12 }