作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 找规律 动态规划 日期 题目地址:https://leetcode.com/problems/divisor-game/ 题目描述 Alice and Bob take turns playing a game, with Alice starting first. Initially, there is a number N on the chal…
题目如下: Alice and Bob take turns playing a game, with Alice starting first. Initially, there is a number N on the chalkboard. On each player's turn, that player makes a move consisting of: Choosing any x with 0 < x < N and N % x == 0. Replacing the n…
Hello everyone, I am a Chinese noob programmer. I have practiced questions on leetcode.com for 2 years. During this time, I studied a lot from many Great Gods' articles. After worship, I always wanted to write an article as they did, and now I take t…
1025. Divisor Game Alice and Bob take turns playing a game, with Alice starting first. Initially, there is a number N on the chalkboard. On each player's turn, that player makes a move consisting of: Choosing any x with 0 < x < N and N % x == 0. Re…
Leetcode之动态规划(DP)专题-1025. 除数博弈(Divisor Game) 爱丽丝和鲍勃一起玩游戏,他们轮流行动.爱丽丝先手开局. 最初,黑板上有一个数字 N .在每个玩家的回合,玩家需要执行以下操作: 选出任一 x,满足 0 < x < N 且 N % x == 0 . 用 N - x 替换黑板上的数字 N . 如果玩家无法执行这些操作,就会输掉游戏. 只有在爱丽丝在游戏中取得胜利时才返回 True,否则返回 false.假设两个玩家都以最佳状态参与游戏. 示例 1: 输入:2…
1071. 字符串的最大公因子 1071. Greatest Common Divisor of Strings 题目描述 对于字符串 S 和 T,只有在 S = T + ... + T(T 与自身连接 1 次或多次)时,我们才认定 "T 能除尽 S". 返回字符串 X,要求满足 X 能除尽 str1 且 X 能除尽 str2. 每日一算法2019/6/17Day 45LeetCode1071. Greatest Common Divisor of Strings 示例 1: 输入:s…
题目如下: Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than o…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 暴力遍历 日期 题目地址:https://leetcode.com/problems/greatest-common-divisor-of-strings/ 题目描述 For strings S and T, we say "T divides S" if and only if S = T + ... + T (T concatenate…
题目如下: For strings S and T, we say "T divides S" if and only if S = T + ... + T (T concatenated with itself 1 or more times) Return the largest string X such that X divides str1 and X divides str2. Example 1: Input: str1 = "ABCABC", st…
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Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 这道题让我们求两数相除,而且规定我们不能用乘法,除法和取余操作,那么我们还可以用另一神器位操作Bit Operation,思路是,如果被除数大于或等于除数,则进行如下循环,定义变量t等于除数,定义计数p,当t的两倍小于等于被除数时,进行如下循环,t扩大一倍,p扩大一倍,然后更…
463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…
博客搬至blog.csgrandeur.com,cnblogs不再更新. 新的题解会更新在新博客:http://blog.csgrandeur.com/2014/01/15/LeetCode-OJ-Solution/ ———————————————————————————————————————— ———————————————————————————————————————— LeetCode OJ 题解 LeetCode OJ is a platform for preparing tech…
You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs. If z liters of water is measurable, you must…
leetcode面试准备:Divide Two Integers 1 题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 接口: public int divide(int dividend, int divisor) 2 思路 题意 不用乘.除.mod 做一个除法运算. 解 直接用除数去一个一个加,直到被除数被超过的话…
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 29: Divide Two Integershttps://oj.leetcode.com/problems/divide-two-integers/ Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT. =…
刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…
一天一道LeetCode系列 (一)题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. (二)解题 这题看起来很简单,一开始想到的方法就是从0开始一次累加除数,一直到比被除数大为止,好无悬念,这样做的结果就是超时了. 用移位来实现除法效率就比较高了.具体思路可以参考二进制除法.下面举个例子来说明. 例如:10/2 即10…
There is a special square room with mirrors on each of the four walls. Except for the southwest corner, there are receptors on each of the remaining corners, numbered 0, 1, and 2. The square room has walls of length p, and a laser ray from the south…
乘风破浪:LeetCode真题_029_Divide Two Integers 一.前言 两个整数相除,不能使用乘法除法和取余运算.那么就只能想想移位运算和加减法运算了. 二.Divide Two Integers 2.1 问题 2.2 分析与解决 通过分析,我们可以想到,如果使用加法,一次次的减下去,每减一次就加一,直到最后的减数小于除数.但是这样的时间复杂度将会是非常的大,比如100000,3这两个数,非常的耗时,那么如何加快运算呢,我们想到了移位运算. public cl…
