Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Note: Division between two integers should truncate toward zero. The given RPN expression…

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Have you met this question in a real interview? Yes Example ["2", "1", &qu…

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Some examples: ["2", "1", "+", "3", "*"] -&g…

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Some examples: ["2", "1", "+", "3", "*"] -&g…

题目: Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are+,-,*,/. Each operand may be an integer or another expression. Some examples: ["2", "1", "+", "3", "*"] -&g…

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Note: Division between two integers should truncate toward zero. The given RPN expression…

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Some examples: 逆波兰表示法计算机,很简单 ,就是一个stack  由符号弹出计算在填入就行了 最后top上的就是计算的最后的结果. class Solution…

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Some examples: ["2", "1", "+", "3", "*"] -&g…

一.题目链接:https://leetcode.com/problems/evaluate-reverse-polish-notation/ 二.题目大意: 给定后缀表达式,求出该表达式的计算结果. 三.题解: 对于这道题目,首先观察后缀表达式(逆波兰表达式)的特点,那就是运算符在操作数的后面,所以每遇到一个运算符,只需找到它之前的最近的两个数字作为操作数,然后求出的结果作为一个新的操作数.很显然,可以使用一个栈来存储操作数,每遇到运算符从栈中取出顶端的两个数运算即可,运算结果再入栈.代码如下:…

原题地址 基本栈操作. 注意数字有可能是负的. 代码: int toInteger(string &s) { ; ] == '-' ? true : false; : ; i < s.length(); i++) { res *= ; res += s[i] - '; } return negative ? -res : res; } int compute(int a, int b, string sign) { ]) { case '+': return a + b; case '-':…