Evaluate the value of an arithmetic expression in Reverse Polish Notation.

最新推荐文章于 2019-06-08 16:38:00 发布

原创最新推荐文章于 2019-06-08 16:38:00 发布 · 766 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #桟

leetcode 专栏收录该内容

18 篇文章

订阅专栏

本文介绍了一种通过栈来解析并计算逆波兰表达式的算法实现。该算法能正确处理加、减、乘、除四种运算符，并支持整数作为操作数。文中提供了一个具体的C++实现示例，包括如何判断运算符、进行数值计算等关键步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

int evalRPN(vector<string> &tokens) { stack<int> operand; for(int i = 0;i < tokens.size(); i++) { if(isOperator(tokens[i])) { char op = tokens[i].at(0); int operand2 = operand.top(); operand.pop(); int operand1 = operand.top(); operand.pop(); switch(op) { case'+': operand.push(operand1 + operand2); break; case'-': operand.push(operand1 - operand2); break; case'*': operand.push(operand1 * operand2); break; case'/': operand.push(operand1 / operand2); break; } } else { operand.push(atoi(tokens[i].c_str())); } } return operand.top(); } bool isOperator(string str) { if(str == "+" || str == "-" || str == "*" || str == "/") return true; return false; } </int></string>