150. Evaluate Reverse Polish Notation (Medium)

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

Solutions

class Solution {

    // Dead simple

    public int evalRPN(String[] tokens) {
        if (tokens == null || tokens.length == 0) {
            return 0;
        }

        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < tokens.length; i++) {
            if (tokens[i].equals("+")) {
                int val1 = stack.pop();
                int val2 = stack.pop();

                stack.push(val2 + val1);

                continue;
            }

            if (tokens[i].equals("-")) {
                int val1 = stack.pop();
                int val2 = stack.pop();

                stack.push(val2 - val1);

                continue;
            }

            if (tokens[i].equals("*")) {
                int val1 = stack.pop();
                int val2 = stack.pop();

                stack.push(val2 * val1);

                continue;
            }

            if (tokens[i].equals("/")) {
                int val1 = stack.pop();
                int val2 = stack.pop();

                stack.push(val2 / val1);

                continue;
            }

            stack.push(Integer.valueOf(tokens[i]));
        }

        return stack.pop();
    }
}

Incorrect Solutions

References

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