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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?
n = 1, return true.
n = 2, return true.
n = 3, return false.
n = 4, return true.
n = 5, return true.
O(n) time and O(1) memory
跟LeetCode上的那道Nim Game几乎一模一样。
解法一：
class Solution { public: /** * @param n: an integer * @return: a boolean which equals to true if the first player will win */ bool firstWillWin(int n) { if (n <= 0) return false; if (n <= 2) return true; vector<bool> dp(n + 1, false); dp[1] = dp[2] = true; dp[3] = false; for (int i = 4; i <= n; ++i) { dp[i] = dp[i - 3]; } return dp.back(); } };
解法二：
class Solution { public: /** * @param n: an integer * @return: a boolean which equals to true if the first player will win */ bool firstWillWin(int n) { return n % 3 != 0; } };
[LintCode] Coins in a Line 一条线上的硬币
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原文地址：http://www.cnblogs.com/grandyang/p/5861500.html
