LeetCode150. Evaluate Reverse Polish Notation

一、题目

You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation.

Evaluate the expression. Return an integer that represents the value of the expression.

Note that:

The valid operators are ‘+’, ‘-’, ‘*’, and ‘/’.
Each operand may be an integer or another expression.
The division between two integers always truncates toward zero.
There will not be any division by zero.
The input represents a valid arithmetic expression in a reverse polish notation.
The answer and all the intermediate calculations can be represented in a 32-bit integer.

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].

二、题解

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<long long> st;
        for(auto token:tokens){
            if(token == "+" || token == "-" || token == "*" || token == "/"){
                long long x1 = st.top();
                st.pop();
                long long x2 = st.top();
                st.pop();
                if(token == "+") st.push(x1 + x2);
                else if(token == "-") st.push(x2 - x1);
                else if(token == "*") st.push(x1 * x2);
                else if(token == "/") st.push(x2 / x1);
            }
            else st.push(stoll(token));
        }
        return st.top();
    }
};