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…

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…

227. 基本计算器 II 227. Basic Calculator II 题目描述 实现一个基本的计算器来计算一个简单的字符串表达式的值. 字符串表达式仅包含非负整数,+,-,*,/ 四种运算符和空格.整数除法仅保留整数部分. LeetCode227. Basic Calculator II中等 示例 1: 输入: "3+2*2" 输出: 7 示例 2: 输入: " 3/2 " 输出: 1 示例 3: 输入: " 3+5 / 2 " 输出:…

[LeetCode]227. Basic Calculator II 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.me/ 题目地址:https://leetcode.com/problems/basic-calculator-ii/description/ 题目描述: Implement a basic calculator to evaluate a simple exp…

这种题都要设置一个符号位的变量 224. Basic Calculator 设置数值和符号两个变量,遇到左括号将数值和符号加进栈中 class Solution { public: int calculate(string s) { stack<int> st; ,flag = ; ;i < s.size();i++){ '){ ; '){ num = num* + flag * (s[i] - '); i++; } res += num; i--; } else if(s[i] ==…

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 should truncate toward zero. You may assu…

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…

Basic Calculator II 题目 思路 和这个一样:Basic Calculator I 代码 class ExpressionTransformation { public: string trans_to_postfix_expression_to_s(string); // 将得到的表达式转化为后缀表达式 long long int calculate_from_postfix_expression(); // 依据后缀表达式计算值 private: vector<string…

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…