Force algorithm buckle title -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 = 

Solution: 
　　use the stack to

 1 class Solution {
 2 public:
 3     int evalRPN(vector<string> &tokens) {
 4         if (tokens.size() == 0)return 0;
 5         stack<int>s;
 6         for (auto a : tokens)
 7         {
 8             if (a == "+" || a == "-" || a == "*" || a == "/")
 9             {
10                 int num2 = s.top();
11                 s.pop();
12                 int num1 = s.top();
13                 s.pop();
14                 int res = 0;
15                 if (a == "+")
16                     res = num1 + num2;
17                 else if (a == "-")
18                     res = num1 - num2;
19                 else if (a == "*")
20                     res = num1 * num2;
21                 else
22                     res = num1 / num2;
23                 s.push(res);
24             }
25             else
26                 s.push(atoi(a.c_str()));
27         }
28         return s.top();
29     }
30 };