问题

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

方法1

stack.

时间复杂度$O(n)$.
空间复杂度$O(n)$.

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> stack = new ArrayDeque<Integer>();
        for (String str : tokens) {
            switch (str) {
                case "+": {
                    int i1 = stack.pop();
                    int i2 = stack.pop();
                    stack.push(i2 + i1);
                    break;
                }
                case "-": {
                    int i1 = stack.pop();
                    int i2 = stack.pop();
                    stack.push(i2 - i1);
                    break;
                }
                case "*": {
                    int i1 = stack.pop();
                    int i2 = stack.pop();
                    stack.push(i2 * i1);
                    break;
                }
                case "/": {
                    int i1 = stack.pop();
                    int i2 = stack.pop();
                    stack.push(i2 / i1);
                    break;
                }
                default:
                    stack.push(Integer.valueOf(str));
            }
        }
        return stack.pop();
    }
}