**[Lintcode]Coins in a Line_there are n coins,each showing either heads or tai-优快云博客

本文链接：https://blog.youkuaiyun.com/jc69186918/article/details/52343928

本文介绍了一种两人轮流从右侧取走一枚或两枚硬币的游戏，并通过数学归纳法确定了先手玩家的胜败策略。给出了具体实现的Java代码示例。

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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?

Have you met this question in a real interview?

Yes

Example

n = 1, return true.

n = 2, return true.

n = 3, return false.

n = 4, return true.

n = 5, return true.

数学归纳法，res[i]代表当一共有i个硬币时,先手是否能赢。当i枚硬币时，因为每次只能取1～2枚，所以当i-1和i-2均为true时，i一定为false

公式总结为:

res[i] = 0 if i <= 2

res[i] = !(res[i - 1] && res[i - 2]) otherwise

public class Solution {
    /**
     * @param n: an integer
     * @return: a boolean which equals to true if the first player will win
     */
    public boolean firstWillWin(int n) {
        boolean[] res = new boolean[n + 1];
        if(n == 0) return false;
        if(n <= 2) return true;
        res[0] = true;
        res[1] = true;
        res[2] = true;
        
        for(int i = 3; i <= n; i++) {
            res[i] = !(res[i - 1] && res[i - 2]);
        }
        return res[n];
    }
}