题目 硬币排成线 有 n 个硬币排成一条线.两个参赛者轮流从右边依次拿走 1 或 2 个硬币,直到没有硬币为止.拿到最后一枚硬币的人获胜. 请判定 第一个玩家 是输还是赢? 样例 n = 1, 返回 true. n = 2, 返回 true. n = 3, 返回 false. n = 4, 返回 true. n = 5, 返回 true. 挑战 O(1) 时间复杂度且O(1) 存储. 解题 两个人在一次拿的时候,当第一个人拿的是1 时,第二个人拿的就是2:当第一个人拿的是2时,第二个人拿的就是1…

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 questi…

题目 硬币排成线 II 有 n 个不同价值的硬币排成一条线.两个参赛者轮流从左边依次拿走 1 或 2 个硬币,直到没有硬币为止.计算两个人分别拿到的硬币总价值,价值高的人获胜. 请判定 第一个玩家 是输还是赢? 样例 给定数组 A = [1,2,2], 返回 true. 给定数组 A = [1,2,4], 返回 false. 解题 动态规划.博弈论 坑死,看了好久 定义dp[i]表示从i到end能取到的最大值 当我们在i处,有两种选择: 1.若取values[i],对方可以取values[i+1…

有 n 个硬币排成一条线.两个参赛者轮流从右边依次拿走 1 或 2 个硬币,直到没有硬币为止.拿到最后一枚硬币的人获胜. 请判定 第一个玩家 是输还是赢? n = 1, 返回 true.n = 2, 返回 true.n = 3, 返回 false.n = 4, 返回 true.n = 5, 返回 true. 解法:DP, 复杂度 O(N), O(N) 最少是n = 3时,返回false,说明当一个player面临只有3个棋子的时候,他肯定会输.dp[i]表示的是,当有i个棋子的时候,先手玩家会不…

There are n coins with different value in a line. Two players take turns to take one or two coins from left side until there are no more coins left. The player who take the coins with the most value wins. Could you please decide the first player will…

输入的n可以分为两种情况: 1. 如果n是3的倍数的话,不论A怎么拿B都可以拿(3-A拿的个数)来使其保持是3的倍数,他就一定能拿到最后一块,所以n是3的倍数的话B必胜 2. 如果n不是3的倍数的话,那么A就能够拿1或者2块使其变为3的倍数,就相当于B变成了第一种情况中的A,所以n不是3的倍数的话A必胜 AC代码: class Solution: """ @param: n: An integer @return: A boolean which equals to true…

https://codesolutiony.wordpress.com/2015/05/24/lintcode-coins-in-a-line-iii/ A very juicy one! Deserve more consideration. class Solution { public: /** * @param values: a vector of integers * @return: a boolean which equals to true if the first playe…

Nice one to learn: DP + Game Theoryhttps://lefttree.gitbooks.io/leetcode/content/dynamicProgramming2/coinsInLine2.html In game theory, we assume the other player always take optimal step too. Combining with DP, we put this assumption together with ea…

Recursion + Memorized Search(DP). And apparently, the code below can be iterative with only 3 vars - DP. class Solution { unordered_map<int, bool> cache; public: bool go(int n) { ) return false; ) return true; if(cache.find(n) == cache.end()) { cach…

Alex and Lee play a game with piles of stones.  There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to end with the most stones.  The total number of stones…