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ylb 提交于 2021-04-21 23:12 . feat: update leetcode problems description

150. Evaluate Reverse Polish Notation

中文文档

Description

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

 

Constraints:

  • 1 <= tokens.length <= 104
  • tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].

Solutions

Python3

import operator

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        opt = {
            "+": operator.add,
            "-": operator.sub,
            "*": operator.mul,
            "/": operator.truediv
        }
        s = []
        for token in tokens:
            if token in opt:
                s.append(int(opt[token](s.pop(-2), s.pop(-1))))
            else:
                s.append(int(token))
        return s[0]

Java

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> s = new ArrayDeque<>();
        int left, right;
        for (String token : tokens) {
            switch(token) {
            case "+":
                right = s.pop();
                left = s.pop();
                s.push(left + right);
                break;
            case "-":
                right = s.pop();
                left = s.pop();
                s.push(left - right);
                break;
            case "*":
                right = s.pop();
                left = s.pop();
                s.push(left * right);
                break;
            case "/":
                right = s.pop();
                left = s.pop();
                s.push(left / right);
                break;
            default:
                s.push(Integer.valueOf(token));
            }
        }
        return s.pop();
    }
}

...

Java
1
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