题目如下: Tic-tac-toe is played by two players A and B on a 3 x 3 grid. Here are the rules of Tic-Tac-Toe: Players take turns placing characters into empty squares (" "). The first player A always places "X" characters, while the second pl…

地址 https://www.acwing.com/solution/LeetCode/content/6670/ 题目描述A 和 B 在一个 3 x 3 的网格上玩井字棋. 井字棋游戏的规则如下: 玩家轮流将棋子放在空方格 (” “) 上.第一个玩家 A 总是用 “X” 作为棋子,而第二个玩家 B 总是用 “O” 作为棋子.“X” 和 “O” 只能放在空方格中,而不能放在已经被占用的方格上.只要有 3 个相同的(非空)棋子排成一条直线(行.列.对角线)时,游戏结束.如果所有方块都放满棋子(不为…

题目如下: Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from either end of the array followed by the player 2 and then player 1 and so on. Each time a player picks a number, that number will not be available f…

[LeetCode]486. Predict the Winner 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/predict-the-winner/description/ 题目描述: Given an array of scores that are non-negative integers…

[292] Nim Game (2019年3月12日,E) 有一堆石头,游戏规则是每次可以从里面拿1-3颗石头,拿到最后的石头的人赢.你和你的对手都 optimal 的玩这个游戏,问先手(也就是你)能不能赢得这个比赛. 题解:我本来写的是 dfs + memo,但是没想到这个题数字太大了.容易爆栈.后来通过看discuss和观察,发现,这个题,只要不是 4 的倍数,先手都能赢得比赛. class Solution { public: bool canWinNim(int n) { != ; }…

[LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 递归和非递归,此提比较简单.广度优先遍历即可.关键之处就在于如何保持访问深度. 下面是4种代码: im…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. Note: Could you optimize your algorithm to use only O(k) extra space? 思路:最简单的方法就是依照[Leetcode]Pascal's Triangle 的方式自顶向下依次求解,但会造成空间的浪费.若仅仅用一个vect…

53. Maximum Subarray[leetcode] Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,1,-5,4],the contiguous subarray [4,-1,2,1] has the largest sum = 6. 挑…

27. Remove Element[leetcode] Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. The order of elements can b…

[刷题][LeetCode]总 用动画的形式呈现解LeetCode题目的思路 参考链接-空 007-整数反转 方法: 弹出和推入数字 & 溢出前进行检查 思路: 我们可以一次构建反转整数的一位数字.在这样做的时候,我们可以预先检查向原整数附加另一位数字是否会导致溢出. 算法: 反转整数的方法可以与反转字符串进行类比. 我们想重复"弹出" x 的最后一位数字,并将它"推入"到 rev 的后面.最后,rev 将与 x 相反. 要在没有辅助堆栈 / 数组的帮助下…