leetcode227. Basic CalculatorII】的更多相关文章

这道题是只有四则运算但是没有括号的,因此可以利用stack来存储的,并且每次使得存储的值与符号有对应的关系,最后在栈中只剩下可以利用加法进行处理的数的,注意在i=n-1的时候到达最后的部分也是需要把数字压入栈中的而不能被忽略掉. class Solution { public: int calculate(string s) { ,num=,n=s.size(); stack<int> st; char op='+'; ;i<n;i++){ '){ num=num*+s[i]-'; }…
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…
一.Basic Calculator Total Accepted: 18480 Total Submissions: 94750 Difficulty: Medium Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, …
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:…
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. Example 1: Input: "3+2*2" Ou…
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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…
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…
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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…
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