力扣算法题—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:
　　使用栈即可

 class Solution {

 public:

     int evalRPN(vector<string> &tokens) {

         if (tokens.size() == )return ;

         stack<int>s;

         for (auto a : tokens)

         {

             if (a == "+" || a == "-" || a == "*" || a == "/")

             {

                 int num2 = s.top();

                 s.pop();

                 int num1 = s.top();

                 s.pop();

                 int res = ;

                 if (a == "+")

                     res = num1 + num2;

                 else if (a == "-")

                     res = num1 - num2;

                 else if (a == "*")

                     res = num1 * num2;

                 else

                     res = num1 / num2;

                 s.push(res);

             }

             else

                 s.push(atoi(a.c_str()));

         }

         return s.top();

     }

 };

巴特西

力扣算法题—150. Evaluate Reverse Polish Notation

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