150. Evaluate Reverse Polish Notation.cpp

/*
Link: 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
*/

// Solution


/*
TC = O(n)
SC = O(n)
*/

class Solution {
public:
    int solve (int x, int y, char z) {          // function for calculations
        switch (z) {
            case '+':
                return x + y;
            case '-':
                return x - y;
            case '*':
                return x * y;
            case '/':
                return x / y;
        }
        return 0;
    }
    
    int evalRPN(vector<string>& tokens) {
        stack<int> st;

        int i, j, len = tokens.size(), x, y;

        for (i = 0; i < len; i++) {
            if (tokens[i] == "+"                        // if the character is an operator
                || tokens[i] == "-" 
                || tokens[i] == "*" 
                || tokens[i] == "/") {
                y = st.top();                           // retrieve the top two numbers
                st.pop();
                x = st.top();
                st.pop();
                st.push (solve(x, y, tokens[i][0]));    // and call the solve function and push it to stack
            } else {
                st.push(stoi(tokens[i]));               // if the character is a number
            }
        }
        return st.top();
    }
    
};