LeetCode:. 150 Reverse Polish expression evaluation

topic:

根据逆波兰表示法,求表达式的值。

有效的运算符包括 +, -, *, / 。每个运算对象可以是整数,也可以是另一个逆波兰表达式。

说明:

整数除法只保留整数部分。
给定逆波兰表达式总是有效的。换句话说,表达式总会得出有效数值且不存在除数为 0 的情况。
示例 1

输入: ["2", "1", "+", "3", "*"]
输出: 9
解释: ((2 + 1) * 3) = 9
示例 2

输入: ["4", "13", "5", "/", "+"]
输出: 6
解释: (4 + (13 / 5)) = 6
示例 3

输入: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
输出: 22
解释: 
  ((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

Source:

class Solution {
    public int evalRPN(String[] tokens) {
        int a = 0;
        int b = 0;
        Stack<Integer> stack = new Stack<Integer>();
        for (String s : tokens) {
            if (s.equals("+")) {
                stack.add(stack.pop() + stack.pop());
            } else if (s.equals("-")) {
                b = stack.pop();
                a = stack.pop();
                stack.add(a - b);
            } else if (s.equals("*")) {
                a = stack.pop();
                b = stack.pop();
                stack.add(a * b);
            } else if (s.equals("/")) {
                b = stack.pop();
                a = stack.pop();
                stack.add(a / b);
            } else {
                stack.add(Integer.valueOf(s));
            }
        }
        return stack.pop();
    }
}
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