这道题是LeetCode里的第991道题. 题目描述: 在显示着数字的坏计算器上,我们可以执行以下两种操作: 双倍(Double):将显示屏上的数字乘 2: 递减(Decrement):将显示屏上的数字减 1 . 最初,计算器显示数字 X. 返回显示数字 Y 所需的最小操作数. 示例 1: 输入:X = 2, Y = 3 输出:2 解释:先进行双倍运算,然后再进行递减运算 {2 -> 4 -> 3}. 示例 2: 输入:X = 5, Y = 8 输出:2 解释:先递减,再双倍 {5 ->…

On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply the number on the display by 2, or; Decrement: Subtract 1 from the number on the display. Initially, the calculator is displaying the num…

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 日期 题目地址:https://leetcode.com/problems/broken-calculator/ 题目描述 On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply t…

On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply the number on the display by 2, or; Decrement: Subtract 1 from the number on the display. Initially, the calculator is displaying the num…

题目如下: On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply the number on the display by 2, or; Decrement: Subtract 1 from the number on the display. Initially, the calculator is displaying t…

On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply the number on the display by 2, or; Decrement: Subtract 1 from the number on the display. Initially, the calculator is displaying the num…

On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply the number on the display by 2, or; Decrement: Subtract 1 from the number on the display. Initially, the calculator is displaying the num…

题目来源:A Broken Calculator 题目如下(链接有可能无法访问): A Broken Calculator Time limit : 2sec / Stack limit : 256MB / Memory limit : 256MB Problem Dave's calculator is broken. His calculator halts when put more than K kinds of number. Dave wants to input an intege…

Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the given expression is…

Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . The expression string contains only non-nega…

Given an expression such as expression = "e + 8 - a + 5" and an evaluation map such as {"e": 1}(given in terms of evalvars = ["e"] and evalints = [1]), return a list of tokens representing the simplified expression, such as […

Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. You may assume that the given ex…

原题链接在这里:https://leetcode.com/problems/basic-calculator-ii/ Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division sh…

原题链接在这里:https://leetcode.com/problems/basic-calculator/ Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and emp…

Basic Calculator Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the giv…

Description: Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. You may assume that…

Description: Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the given e…

978. Basic Calculator https://www.lintcode.com/problem/basic-calculator/description public class Solution { /** * @param s: the given expression * @return: the result of expression */ public int calculate(String s) { // Write your code here Stack<Int…

时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:1821 解决:528 题目描述: Today, facing the rapid development of business, SJTU recognizes that more powerful calculator should be studied, developed and appeared in future market shortly. SJTU now invites you attending such amaz…

1004: 1.1.4Broken Necklace 坏掉的项链 时间限制: 1 Sec  内存限制: 128 MB提交: 11  解决: 9[提交] [状态] [讨论版] [命题人:外部导入] 题目描述 1.1.4Broken Necklace 坏掉的项链 (beads.pas/c/cpp) 题目描述 你有一条由N个红色的,白色的,或蓝色的珠子组成的项链(3<=N<=350),珠子是随意安排的. 这里是 n=29 的二个例子: 1 2 1 2 r b b r b r r b r b b b…

class Solution {public:    int calculate(string s) {        stack<int> num;        stack<char> symbol;        for(int i=0;i<s.length();i++){            if(s[i]==' ')   continue;            else if(s[i]=='('||s[i]=='+'||s[i]=='-')  symbol.pu…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step: Copy All: You can copy all the characters present on the notepad (partial copy is not allowed). Paste: You can paste the character…

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step: Copy All: You can copy all the characters present on the notepad (partial copy is not allowed). Paste: You can paste the character…

本文总结LeetCode上有数学类的算法题,推荐刷题总数为40道.具体考点分析如下图: 1.基本运算问题 题号:29. 两数相除,难度中等 题号:166. 分数到小数,难度中等 题号:372. 超级次方,难度中等 题号:483. 最小好进制,难度困难 题号:810. 黑板异或游戏,难度困难 2.组合数学问题(排列问题) 题号:60. 第k个排列,难度中等 题号:233. 数字 1 的个数,难度困难 题号:670. 最大交换,难度中等 题号:1012. 至少有 1 位重复的数字,难度困难 3.质数…

使用qt designer ,按装anaconda后,在如下路径找到: conda3.05\Library\bin designer.exe文件,双击启动: ​ 创建窗体,命名为XiaoDing,整个的界面如下所示: ​ qt 设计器提供的常用控件基本都能满足开发需求,通过拖动左侧的控件,很便捷的就能搭建出如下的UI界面,比传统的手写控件代码要方便很多. 最终设计的计算器XiaoDing界面如下, ​ 比如,其中一个用于计算器显示的对象:lcdNumber,对象的类型为:LCD Number.右…

微信小程序开发计算器有多种方法,但是大部分代码比较复杂.不容易理解.本案例进行了改进,主要是组件bindtap属性绑定的自定义函数clickBtn(),采用了switch语句,使得代码结构更加清晰,学习者更容易理解,现分享如下.案例的效果如下图所示: 制作步骤如下: 1. 在微信开发者工具的全局控制app.json文件pages数组中输入“pages/calculator/calcu”,建立计算器页面相关文件,在window属性中更改导航标题为“简单计算器”.app.json的代码如下: { "…

题目链接: http://code.google.com/codejam/contest/5214486/dashboard Problem A. Minesweeper 题目意思: 扫雷.告诉地雷所在的位置,求至少须要几步,把全部的非雷位置都翻过来.假设某一个格子周围都不是地雷.则翻转该格子会把周围的格子都翻转过来. 解题思路: dfs找连通块 先把非雷格子周围的地雷总数求出,然后对于周围没有地雷的格子找连通块.剩下的每一个格子都须要再翻转一次. 代码: //#include<CSpreadS…

贪心基础 贪心(Greedy)常用于解决最优问题,以期通过某种策略获得一系列局部最优解.从而求得整体最优解. 贪心从局部最优角度考虑,只适用于具备无后效性的问题,即某个状态以前的过程不影响以后的状态.紧接下来的状态仅与当前状态有关.和分治.动态规划一样,贪心是一种思路,不是解决某类问题的具体方法. 应用贪心的关键,是甄别问题是否具备无后效性.找到获得局部最优的策略.有的问题比较浅显,例如一道找零钱的题目 LeetCode 860. Lemonade Change: // 860. Lemonad…

1:系统必要进程system process    进程文件: [system process] or [system process]进程名称: Windows内存处理系统进程描述: Windows页面内存管理进程,拥有0级优先.alg.exe       进程文件:alg or alg.exe 进程名称:应用层网关服务  描述:这是一个应用层网关服务用于网络共享csrss.exe      进程文件:csrss or csrss.exe 进程名称:Client/Server Runtime…