2. Evaluate Reverse Polish Notation 150

150. Evaluate Reverse Polish Notation

Medium

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, and /. Each operand may be an integer or another expression.

Note that division between two integers should truncate toward zero.

It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, and there will not be any division by zero operation.

 

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].

Accepted

443,625

Submissions

1,025,826

Solution :

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. For example: 

["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 

["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

Approach - Using Stack:

1. Take a Stack and Operators String containing all the operators.
2. Check if token is not in operators push into stack
3. Else store two pop() and switch using token case according to operators

4. Return the pop() in Integer.

Code: 

class Solution {
    public int evalRPN(String[] tokens) {
        Stack<String> st = new Stack<>();
        String op = "+-*/";
        
        for(String t : tokens) {
            if(!op.contains(t)) {
                st.push(t);
            }
            else {
                int a = Integer.valueOf(st.pop());
                int b = Integer.valueOf(st.pop());
                switch(t) {
                    case "+":
                        st.push(String.valueOf(a+b));
                        break;
                    case "-":
                        st.push(String.valueOf(b-a));
                        break;
                    case "*":
                        st.push(String.valueOf(a*b));
                        break;
                    case "/":
                        st.push(String.valueOf(b/a));
                        break;
                }
            }
        }
        return Integer.valueOf(st.pop());
    }
}

Space is O(n) and time is O(n).