LeetCode Top Interview Questions 150. Evaluate Reverse Polish Notation (Java版; Medium)

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

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

/*
乍一看是递归的感觉, 又感觉不太好实现, 然后思考栈的可行性, 发现贼合适
细节: str==与str.equals()
*/
class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> stack = new Stack<>();
        for(int i=0; i<tokens.length; i++){
            String cur = tokens[i];
            if(cur.equals("+") || cur.equals("-") || cur.equals("*") || cur.equals("/")){
                int b = stack.pop();
                int a = stack.pop();
                if(cur.equals("+"))
                    stack.push(a+b);
                else if(cur.equals("-"))
                    stack.push(a-b);
                else if(cur.equals("*"))
                    stack.push(a*b);
                else
                    stack.push(a/b);
            }
            else{
                stack.push(Integer.parseInt(cur));
            }
        }
        return stack.peek();
    }
}