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

A:

       每遇到运算符，就提出最后面的两个，运算，并保存。

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> mystack;
        for(auto s:tokens)
        {
            if(s == "+" || s == "-" || s == "*" || s == "/")
            {
                int a = mystack.top();
                mystack.pop();
                int b = mystack.top();
                mystack.pop();
                if(s == "+")
                    mystack.push(b+a);
                if(s == "-")
                    mystack.push(b-a);
                if(s == "*")
                    mystack.push(b*a);
                if(s == "/")
                    mystack.push(b/a);
            }else{
                mystack.push(stoi(s));
            }
        }
        return mystack.top();
    }
};

Mistakes:
1 : 没有考虑 当一个数为：负数的时候， 例如 -2，  如果只检查首字母， 就会误认为是运算符。
2: 刚开始，自以为是了。  用double类型保持中间值。
    没想到，作者是不用的。   用int数组就行。

----------第二遍-----------
3： 由于是从 + 的情况直接copy 的代码，  对于 -   / 的情况，没有考虑 a，b 的顺序要调过来的。


Learned: