877. Stone Game

本文介绍了一个涉及策略选择的游戏——石头游戏。在这个游戏中,两名玩家轮流从一堆石头中取走最左边或最右边的一堆,目标是在游戏结束时拥有最多的石头。文章通过递归算法详细解释了如何确定先手玩家是否能赢得比赛。

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Alex and Lee play a game with piles of stones.  There are an even 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.  The total number of stones is odd, so there are no ties.

Alex and Lee take turns, with Alex starting first.  Each turn, a player takes the entire pile of stones from either the beginning or the end of the row.  This continues until there are no more piles left, at which point the person with the most stones wins.

Assuming Alex and Lee play optimally, return True if and only if Alex wins the game.

 

Example 1:

Input: [5,3,4,5]
Output: true
Explanation: 
Alex starts first, and can only take the first 5 or the last 5.
Say he takes the first 5, so that the row becomes [3, 4, 5].
If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alex, so we return true.

 

Note:

  1. 2 <= piles.length <= 500
  2. piles.length is even.
  3. 1 <= piles[i] <= 500
  4. sum(piles) is odd.

 

min max search

class Solution:
    def stoneGame(self, a):
        """
        :type piles: List[int]
        :rtype: bool
        """
        su = sum(a)
        ma,mi = {},{}
        def getmax(i,j):
            if i==j: return a[i]
            if (i,j) in ma: return ma[(i,j)]
            ma[(i,j)] = max(a[i]+getmin(i+1,j), a[j]+getmin(i,j-1))
            return ma[(i,j)]
        def getmin(i,j):
            if i==j: return 0
            if (i,j) in mi: return mi[(i,j)]
            mi[(i,j)] = min(getmax(i+1,j), getmax(i,j-1))
            return mi[(i,j)]
        
        return  getmax(0,len(a)-1)>su//2
    

            

 

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