leetcode [150]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

题目大意：

计算用 Reverse Polish Notation 表达的算术表达式的值。

解法：

题目有说明给定的表达式一定是一个有效的表达式，采用一个栈来保存数字，一旦遇到操作符，从栈中弹出数字，进行计算，然后再保存到栈中。

java:

class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer>s=new Stack<Integer>();
        for(String str:tokens){
            if(str.equals("+")){
                int n1=s.pop();
                int n2=s.pop();
                s.add(n1+n2);
            }else if(str.equals("-")){
                int n1=s.pop();
                int n2=s.pop();
                s.add(n2-n1);
            }else if(str.equals("*")){
                int n1=s.pop();
                int n2=s.pop();
                s.add(n1*n2);
            }else if(str.equals("/")){
                int n1=s.pop();
                int n2=s.pop();
                s.add(n2/n1);
            }else{
                s.add(Integer.parseInt(str));
            }
        }
        return s.pop();
    }
}