LintCode Coins in a Line

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:

public class Solution {
    public boolean firstWillWin(int n) {
        int [] dp = new int[n+1];
        return memorySearch(n, dp);
    }
    
    private boolean memorySearch(int n, int [] dp){
        if(dp[n] != 0){
            return dp[n] == 2;
        }
        
        if(n<=0 || n == 3){
            dp[n] = 1;
        }else if(n == 1 || n == 2){
            dp[n] = 2;
        }else{
            if((memorySearch(n-2, dp) && memorySearch(n-3, dp)) 
                || (memorySearch(n-3, dp) && memorySearch(n-4, dp))){
                    dp[n] = 2;
                }else{
                    dp[n] = 1;
                }
        }

        if(dp[n] == 2){
            return true;
        }else{
            return false;
        }
    }
}

 跟上Coins in a Line II.