Coins in a Line

Posted Sheryl Wang

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Coins in a Line相关的知识,希望对你有一定的参考价值。

There are n coins in a line. Two players take turns to take one or two coins from right side until there are no more coins left. The player who take the last coin wins.

Could you please decide the first play will win or lose?

这是一类博弈类的DP问题.

博弈问题的基本解决思路是:

state:定义其中一个人的状态.

function:考虑两个人的状态做更新,即根据相互对抗的一个关系来做决策.

博弈类题目适合从大的状态往小的状态递推,非常适合记忆化搜索.不过这题本身比较简单,可以已经bottom-up来做.

这里分析f[i]为还剩i枚硬币的时候先手能否获胜.我们考虑递推状态.先手在f[i]可以获胜,即他拿了最后一枚或者两枚硬币.这样在f[i-1]后者f[i-2]时是后手获胜的,所以有递推关系f[i] = !f[i-1]||!f[i-2]

初始化,还剩0枚硬币时先手肯定输.还剩1,2枚硬币时先手能赢.但是最后的结果取什么呢.是还剩0枚硬币,还是n枚.其实我们是从还有n枚硬币,考虑这时候的博弈状态考量的,所以是f[n].代码如下:

class Solution:
    # @param n: an integer
    # @return: a boolean which equals to True if the first player will win
    def firstWillWin(self, n):
        if n <= 0:
            return False
        if n <= 2:
            return True
        f = [False] *(n+1)
        f[1] = True
        f[2] = True
        
        for i in xrange(2,n+1):
            f[i] = (not f[i-1]) or (not f[i-2])
            
        return f[n]

 

 

以上是关于Coins in a Line的主要内容,如果未能解决你的问题,请参考以下文章

Coins in a Line

Coins in a Line II

Coins in a Line III

[LintCode] Coins in a Line II

LeetCode Coins in a Line

Coins in a Line II