逆波兰表达式求值

题目 LeetCode (opens new window)

根据 逆波兰表示法，求表达式的值。

有效的算符包括 +、-、*、/ 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。

注意 两个整数之间的除法只保留整数部分。

可以保证给定的逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。

示例 1：

输入：tokens = ["2","1","+","3","*"]
输出：9
解释：该算式转化为常见的中缀算术表达式为：((2 + 1) * 3) = 9

示例 2：

输入：tokens = ["4","13","5","/","+"]
输出：6
解释：该算式转化为常见的中缀算术表达式为：(4 + (13 / 5)) = 6

示例 3：

输入：tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
输出：22
解释：该算式转化为常见的中缀算术表达式为：
  ((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

代码

/**
 * @param {string[]} tokens
 * @return {number}
 */
var evalRPN = function (tokens) {
    const len = tokens.length;
    const stack = new MyStack();
    for (let i = 0; i < len; i++) {
        if (tokens[i] == "+") {
            let rNum = stack.pop();
            let lNum = stack.pop();
            let res = lNum + rNum;
            stack.push(res);
        } else if (tokens[i] == "-") {
            let rNum = stack.pop();
            let lNum = stack.pop();
            let res = lNum - rNum;
            stack.push(res);
        } else if (tokens[i] == "*") {
            let rNum = stack.pop();
            let lNum = stack.pop();
            let res = lNum * rNum;
            stack.push(res);
        } else if (tokens[i] == "/") {
            let rNum = stack.pop();
            let lNum = stack.pop();
            let res = lNum / rNum;
            if (res < 0) {
                stack.push(Math.ceil(res));
            } else {
                stack.push(Math.floor(res));
            }
        } else { // 是数字，入栈
            stack.push(+tokens[i])
        }
    }
    return stack.pop();
};