LeetCode 150. Evaluate Reverse Polish Notation

原题链接在这里：https://leetcode.com/problems/evaluate-reverse-polish-notation/

题目：

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

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

Some examples: 

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

题解：

利用stack, 遇到数字就压栈，遇到运算符就先pop() op2, 再pop() op1, 按op1 运算符op2 计算，得出结果压回栈，最后站内剩下的就是结果.

Time Complexity: O(n). n = tokens.length.

Space: O(n).

AC Java:

class Solution {
    public int evalRPN(String[] tokens) {
        if(tokens == null || tokens.length == 0){
            return 0;
        }
        
        Stack<Integer> stk = new Stack<Integer>();
        for(int i = 0; i<tokens.length; i++){
            if(tokens[i].equals("+")){
                int op2 = stk.pop();
                int op1 = stk.pop();
                stk.push(op1+op2);
            }else if(tokens[i].equals("-")){
                int op2 = stk.pop();
                int op1 = stk.pop();
                stk.push(op1-op2);
            }else if(tokens[i].equals("*")){
                int op2 = stk.pop();
                int op1 = stk.pop();
                stk.push(op1*op2);
            }else if(tokens[i].equals("/")){
                int op2 = stk.pop();
                int op1 = stk.pop();
                stk.push(op1/op2);
            }else{
                stk.push(Integer.valueOf(tokens[i]));
            }
        }
        
        return stk.pop();
    }
}

跟上Basic Calculator.