LeetCode 1140. Stone Game II解题报告(python)_1140. stone game ii python-优快云博客

本文链接：https://blog.youkuaiyun.com/Orientliu96/article/details/103542669

本文探讨了StoneGameII游戏中的最优策略，通过动态规划算法解决Alex和Lee的石头游戏问题，旨在帮助玩家理解如何在限定条件下获得最大石子数。

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1140. Stone Game II

Stone Game II python solution

题目描述

Alex and Lee continue their games with piles of stones. There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to end with the most stones.
Alex and Lee take turns, with Alex starting first. Initially, M = 1.
On each player’s turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M. Then, we set M = max(M, X).
The game continues until all the stones have been taken.
Assuming Alex and Lee play optimally, return the maximum number of stones Alex can get.
在这里插入图片描述

解析

使用动态规划，使用A[i]表示piles[i:]中以[1,2*M]为取值范围可以取的最多石子数
动态规划的方程如下所示
$f (i, m) = A [i] - m i n (f (i + x, m a x (M, x)))$
如果想让亚历克斯拿的尽可能多，那么我们就应该让李少拿。亚历克斯拿到最多的石子个数是，整体的石子个数减去李拿最少的石子，即min(f(i+x,max(M, x)))。

from functools import lru_cache
class Solution:
    def stoneGameII(self, A: List[int]) -> int:
        N = len(A)
        for i in range(N - 2, -1, -1):
            A[i] += A[i + 1]
        from functools import lru_cache
        @lru_cache(None)
        def dp(i, m):
            if i + 2 * m >= N: return A[i]
            return A[i] - min(dp(i + x, max(m, x)) for x in range(1, 2 * m + 1))
        return dp(0, 1)