LeetCode - 150. Evaluate Reverse Polish Notation

1. 问题描述

• 描述：逆波兰表达式求值
• 输入：逆波兰表达式数组
• 输出：最后的结果

2. 解题思路

• 关键点：如果当前字符为变量或者为数字，则压栈，如果是运算符，则将栈顶两个元素弹出作相应运算，结果再入栈，最后当表达式扫描完后，栈里的就是结果。
• 解题步骤：
  • a. 构建栈
  • b. 求解
• javascript

/**
 * @param {string[]} tokens
 * @return {number}
 */
var evalRPN = function (tokens) {
    let stack = []
    for (let token of tokens) {
        let t = Number(token);
        if (isNaN(t)) {
            let n1 = stack.pop();
            let n2 = stack.pop();
            switch (token) {
                case '+':
                    stack.push(n2 + n1);
                    break;
                case '-':
                    stack.push(n2 - n1);
                    break;
                case '*':
                    stack.push(n2 * n1);
                    break;
                case '/':
                    let d = n2 / n1;
                    if (d < 0) {
                        d = - Math.floor(-d);
                    }else{
                        d = Math.floor(d);
                    }
                    stack.push(d);
                    break;
            }
        } else {
            stack.push(t);
        }
    }
    return stack[0];
};

• python

class Solution:
    def evalRPN(self, tokens) -> int:
        stack = []
        for token in tokens:
            if token == '+':
                n1 = stack.pop()
                n2 = stack.pop()
                stack.append(n2 + n1)
            elif token == '-':
                n1 = stack.pop()
                n2 = stack.pop()
                stack.append(n2 - n1)
            elif token == '*':
                n1 = stack.pop()
                n2 = stack.pop()
                stack.append(n2 * n1)
            elif token == '/':
                n1 = stack.pop()
                n2 = stack.pop()
                if n2 // n1 < 0:
                    stack.append(-(abs(n2) // abs(n1)))
                else:
                    stack.append(n2 // n1)
            else:
                stack.append(int(token))
            
        return stack[0]

3. 小细节

• 整除，题目中除法为整除，且要注意负数整除。