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

逆波蘭表示式求值
根據逆波蘭表示法，求表示式的值。

有效的運算子包括 +, -, *, / 。每個運算物件可以是整數，也可以是另一個逆波蘭表示式。

說明：

整數除法只保留整數部分。
給定逆波蘭表示式總是有效的。換句話說，表示式總會得出有效數值且不存在除數為 0 的情況。
示例 1：

輸入: [“2”, “1”, “+”, “3”, “*”]
輸出: 9
解釋: ((2 + 1) * 3) = 9
示例 2：

輸入: [“4”, “13”, “5”, “/”, “+”]
輸出: 6
解釋: (4 + (13 / 5)) = 6
示例 3：

輸入: [“10”, “6”, “9”, “3”, “+”, “-11”, “", “/”, "

這題雖然是Medium水平，但很簡單的，easy系列水平。
本來是想用switch case語句 case裡不能是string ;所以此題用的if ==
因為定義的棧是整形，所以不能直接入棧。藉助stoic()。
class Solution {
public:
int evalRPN( vector& tk ) {
int sum = 0;
stack < int > st ;
for ( int i = 0 ; i < tk.size() ; i++){

        if (tk[i] == "+"||tk[i] == "-"||tk[i] == "*"||tk[i] == "/")
           
        { 
          int  s1 = st.top() ; 
              st.pop() ; 
            int  s2 = st.top();
             st.pop() ;
            if(tk[i] == "+" )                                     
                    st.push( s1 + s2 );
                   
              else if (tk[i] == "-" )
                    st.push( s2 - s1 );
                   
                   else if (tk[i] == "*" )
                     st.push( s1 * s2 );
                      else if  (tk[i] == "/" )
                           st.push( s2 / s1 );                                                                                
        }            
   else   st.push ( stoi(tk[i]) );    
}
    return st.top() ;
        }

};