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", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 之前已经讲过逆波兰表达式的生成和运算方法，实际上计算是更为简单的。具体见博客。 package com.liuhao.acm.leetcode;import java.util...
Total Accepted: 55722 Total Submissions: 249668 Difficulty: Medium 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", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 (M) Basic Calculator (H) Expression Add Operators class Solu...
初看貌似有点复杂，可是搞懂了很easy，就一个简单的栈应用，每次遇到计算符号"+", "-", "*", "/"就将栈顶端两个数字出栈，计算后再将结果压栈就可以。。#include<iostream> #include<vector> #include<stack> using namespace std;class Solution { public:int evalRPN(vector<string> &tokens) {stack<int> stack;for (int i = 0; i < tokens.size(); i++){if (tokens[i] == "*" || tokens[i] == "-" || tokens[i] == "+" || toke...
求在 逆波兰表示法 中算术表达式的值。有效的运算符号包括 +, -, *, / 。每个运算对象可以是整数，也可以是另一个逆波兰计数表达。例如： ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6详见：https://leetcode.com/problems/evaluate-reverse-polish-notation/description/class Solution { public:int evalRPN(vector<string> &tokens) {if(tokens.size() == 0)return 0...
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", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6The concept is:When meeting the number, push into the stack.When meeting the operator, pop the top 2 number and compute the val...
仅提供个人的一种解题思路，未必是最优，仅供各位参考！import java.util.Stack;/*** * <p>* ClassName SolutionEvaluateReversePolishNotation* </p>* <p>* Description 该题是解逆波兰表达式 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", "*"] -> ((2 + 1) ...
问题描述：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 is always valid. That means the expression would always evaluate to a result and there won‘t be any divide by zero operation.Example 1:Input: ...
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", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 就是求逆波兰表达式（后续遍历）的结果。1、直接求解，很慢publicclass Solution {publicint evalRPN(String[] tokens) {int len = tok...
初看貌似有点复杂，但是搞懂了非常简单，就一个简单的栈应用，每次遇到计算符号"+", "-", "*", "/"就将栈顶端两个数字出栈，计算后再将结果压栈即可。。#include<iostream> #include<vector> #include<stack> using namespace std;class Solution { public:int evalRPN(vector<string> &tokens) {stack<int> stack;for (int i = 0; i < tokens.size(); i++){if (tokens[i] == "*" || tokens[i] == "-" || tokens[i] == "+" || toke...
【151-Evaluate Reverse Polish Notation（计算逆波兰式）】【LeetCode-面试算法经典-Java实现】【所有题目目录索引】原题 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", "*"] ->((2 + 1) * 3) ->9["4", "13", "5", "/", "+"] ->(4 + (13 / 5)) ->6题目大意 ...
Valid operators are?+,?-,?*,?/. Each operand may be an integer or another expression.public class Solution {public int evalRPN(String[] tokens) {int result=0;List<Integer> arrayList=new LinkedList<Integer>();int temp1=0;int temp2=0;for(String temp:tokens){if(temp.compareTo("+")==0){temp1=arrayList.remove(arrayList.size()-1);temp2=arrayList.remove(arrayList.size()-1);result=temp2+temp1;arrayList.ad...
前言昨天终于项目完成，历时一个多月的时间，解决了Android上内存泄露和各种ANR问题，今天终于有时间写点ACM题目了。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:思路这道题目还是挺奇怪的，读完题目赶紧非常简单，但是LeetCode上的AC率只有19%，尝试第一次提交代码就ac了，有点无语，...
问题描述：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", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6原问题链接：https://leetcode.com/problems/evaluate-reverse-polish-notation/问题分析 关于逆波兰表达式的计算，其实是有...
遇到运算符将栈底两个元素弹出，进行计算class Solution { public:int evalRPN(vector<string>& tokens) {stack<int> stk;int n = tokens.size();for (int i = 0; i < n; i++) {string& token = tokens[i];if (isNumber(token)) {stk.push(atoi(token.c_str()));} else {int num2 = stk.top();stk.pop();int num1 = stk.top();stk.pop();switch (token[0]) {case ‘+‘:stk.push(num1 + num2);break;case ‘-‘:stk.push(num1 - nu...
【LeetCode】150. Evaluate Reverse Polish Notation 逆波兰表达式求值（Medium）（JAVA） 题目地址: https://leetcode.com/problems/evaluate-reverse-polish-notation/ 题目描述： 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 ...