[LeetCode] Problem 150 - Evaluate Reverse Polish Notation

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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

No.1

Input: [“2”, “1”, “+”, “3”, “*”]

Output: 9

Explanation: ((2 + 1) * 3) = 9

No.2

Input: [“4”, “13”, “5”, “/“, “+”]

Output: 6

Explanation: (4 + (13 / 5)) = 6

No.3

Input: [“10”, “6”, “9”, “3”, “+”, “-11”, “*“, “/“, “*”, “17”, “+”, “5”, “+”]

Output: 22

Explanation: 

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  ((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

Code

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public int evalRPN(String[] tokens) {
    if (tokens == null || tokens.length < 1)
        return 0;
    
    Stack<Integer> stack = new Stack<>();
    Set<String> operation = new HashSet<String>(){{
        add("+");
        add("-");
        add("*");
        add("/");
    }};

    for (String token : tokens) {
        if (!operation.contains(token)) {
            stack.push(Integer.valueOf(token));
            continue;
        }

        int b = stack.pop();
        int a = stack.pop();

        switch (token) {
            case "+" :
                stack.push(a + b);
                continue;
            case "-" :
                stack.push(a - b);
                continue;
            case "*" :
                stack.push(a * b);
                continue;
            case "/" :
                stack.push(a / b);
                continue;
        }
    }

    return stack.peek();
}