150. Evaluate Reverse Polish Notation

最新推荐文章于 2025-08-17 21:38:20 发布

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最新推荐文章于 2025-08-17 21:38:20 发布

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文章标签： java runtime

原文链接：https://segmentfault.com/a/1190000018171255

本文介绍了一种使用栈解决反波兰表达式计算问题的方法。通过分析输入的字符串数组，运用栈来存储数值并进行相应的运算，实现了对复杂算术表达式的有效解析。文章提供了具体的代码实现，展示了如何处理加减乘除四种基本运算。

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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
= 22

难度：medium

题目：
求算术表达式在反波兰表示法中的值。
有效的操作符是+、-、*、/。每个操作数可以是一个整数或另一个表达式。

思路：栈

Runtime: 6 ms, faster than 91.00% of Java online submissions for Evaluate Reverse Polish Notation.
Memory Usage: 36.6 MB, less than 100.00% of Java online submissions for Evaluate Reverse Polish Notation.

class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> stack = new Stack<>();
        int tNum = 0;
        for (int i = 0; i < tokens.length; i++) {
            switch(tokens[i]) {
                case "+" :
                    stack.push(stack.pop() + stack.pop());
                    break;
                case "-" :
                    tNum = stack.pop();
                    stack.push(stack.pop() - tNum);
                    break;
                case "*" :
                    stack.push(stack.pop() * stack.pop());
                    break;
                case "/" :
                    tNum = stack.pop();
                    stack.push(stack.pop() / tNum);
                    break;
                default :
                    stack.push(Integer.parseInt(tokens[i]));
            }
        }
        
        return stack.pop();
    }
}