#Leetcode# 150. Evaluate Reverse Polish Notation

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

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

代码：

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<string> stck;
        for (auto op : tokens) {
            if (op == "+") {
                int operand2 = stoi(stck.top()); stck.pop();
                int operand1 = stoi(stck.top()); stck.pop();
                int result = operand1 + operand2;
                stck.push(to_string(result));
            } else if (op == "-") {
                int operand2 = stoi(stck.top()); stck.pop();
                int operand1 = stoi(stck.top()); stck.pop();
                int result = operand1 - operand2;
                stck.push(to_string(result));
            } else if (op == "*") {
                int operand2 = stoi(stck.top()); stck.pop();
                int operand1 = stoi(stck.top()); stck.pop();
                int result = operand1 * operand2;
                stck.push(to_string(result));
            } else if (op == "/") {
                int operand2 = stoi(stck.top()); stck.pop();
                int operand1 = stoi(stck.top()); stck.pop();
                int result = operand1 / operand2;
                stck.push(to_string(result));
            } else {
                stck.push(op);
            }
        }
        int res = stoi(stck.top()); stck.pop();
        return res;
    }
};

大一的时候记得讲过这个东西 用到 $stack$ 遇到符号的时候从栈里弹出两个进行运算然后把结果再压进去