[刷题防痴呆] 0150 - 逆波兰表达式求值 (Evaluat

题目地址

https://leetcode.com/problems/evaluate-reverse-polish-notation/

题目描述

150. Evaluate Reverse Polish Notation

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

Valid operators are +, -, *, and /. Each operand may be an integer or another expression.

Note that division between two integers should truncate toward zero.

It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, and there will not be any division by zero operation.

 

Example 1:

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

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

Input: tokens = ["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

思路

• Stack.
• 遇到数字就进栈. 遇到运算符就弹出两个数进行运算, 结果再进栈.

关键点

代码

• 语言支持：Java

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> stack = new ArrayDeque<>();
        for (String str: tokens) {
            if (str.equals("+")) {
                int a = Integer.valueOf(stack.pop());
                int b = Integer.valueOf(stack.pop());
                stack.push(b + a);
            } else if (str.equals("-")) {
                int a = Integer.valueOf(stack.pop());
                int b = Integer.valueOf(stack.pop());
                stack.push(b - a);
            } else if (str.equals("*")) {
                int a = Integer.valueOf(stack.pop());
                int b = Integer.valueOf(stack.pop());
                stack.push(b * a);
            } else if (str.equals("/")) {
                int a = Integer.valueOf(stack.pop());
                int b = Integer.valueOf(stack.pop());
                stack.push(b / a);
            } else {
                stack.push(Integer.valueOf(str));
            }
        }

        return stack.pop();
    }
}