leetcode150 逆波兰表达式求值-优快云博客

本文链接：https://blog.youkuaiyun.com/SmallNancy/article/details/99815980

本文介绍了一种使用栈来解决逆波兰表达式求值问题的方法。通过解析输入的逆波兰表达式，利用栈的数据结构特性，实现了加、减、乘、除四种运算的高效处理。文章提供了一个Java实现的例子，展示了如何读取输入、解析表达式并最终得到结果。

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.

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: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9
Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation: 
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5

思路：用一个栈便可以解决。

package com.leetcode;

import java.util.Scanner;
import java.util.Stack;

/**
 * @author YanQiKing
 * @date 2019/8/20 8:17
 * leetcode 150 逆波兰表达式求值
 */
public class ReversePolish {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine().trim();
        String[] tokens = s.split(" ");
        System.out.println(evalRPN(tokens));
    }
    public static int evalRPN(String[] tokens) {
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < tokens.length;i++) {
            if (tokens[i].equals("+") || tokens[i].equals("-") || tokens[i].equals("*") || tokens[i].equals("/")) {
                int a = stack.pop();
                int b = stack.pop();
                if (tokens[i].equals("+") ) {
                   stack.add(a+b);
                }else if (tokens[i].equals("-") ){
                    stack.add(b - a);
                }else if (tokens[i].equals("*")) {
                    stack.add(b * a);
                }else {
                    stack.add(b /a);
                }
            }else {
                stack.add(Integer.parseInt(tokens[i]));
            }
        }
        return stack.peek();

    }
}