LintCode Coins in a Line

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 119 阅读

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原文链接：http://www.cnblogs.com/Dylan-Java-NYC/p/6406922.html

文章标签：

#java

本文探讨了一个经典的博弈论问题——取币游戏。通过动态规划的方法解决了两人轮流从一排硬币中取走一枚或两枚的问题，目的是决定先手玩家是否会赢得比赛。文章提供了详细的算法思路和Java实现。

原题链接在这里：http://www.lintcode.com/en/problem/coins-in-a-line/

题目：

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?

Example

n = 1, return true.

n = 2, return true.

n = 3, return false.

n = 4, return true.

n = 5, return true.

题解：

类似Nim Game. 这道题其实是DP, 后面还有进阶题.

状态还剩i个coin时先手的人是赢是输.

转移方程 dp[i] = (dp[i-2] && dp[i-3]) || (dp[i-3] && dp[i-4]).

初始化 dp[0] = false, dp[1] = true, dp[2] = true, dp[3] = false.

答案dp[n].

Time Complexity: O(n), 因为中间用了dp来记录中间值. Space: O(n), stack space.

AC Java:

 1 public class Solution {
 2     public boolean firstWillWin(int n) {
 3         int [] dp = new int[n+1];
 4         return memorySearch(n, dp);
 5     }
 6     
 7     private boolean memorySearch(int n, int [] dp){
 8         if(dp[n] != 0){
 9             return dp[n] == 2;
10         }
11         
12         if(n<=0 || n == 3){
13             dp[n] = 1;
14         }else if(n == 1 || n == 2){
15             dp[n] = 2;
16         }else{
17             if((memorySearch(n-2, dp) && memorySearch(n-3, dp)) 
18                 || (memorySearch(n-3, dp) && memorySearch(n-4, dp))){
19                     dp[n] = 2;
20                 }else{
21                     dp[n] = 1;
22                 }
23         }
24 
25         if(dp[n] == 2){
26             return true;
27         }else{
28             return false;
29         }
30     }
31 }