Evaluate Reverse Polish Notation

 1 class Solution {
 2     stack<int>str;
 3 public:
 4     int evalRPN(vector<string>& tokens) {
 5         string c;
 6         int opa,opb;
 7         for(auto c:tokens)
 8         {
 9            if(c.size()>1||isdigit(c[0]))str.push(stoi(c));  //isdigit()是把字符读取后变成整数再判断
10         else 
11         {
12             opb=str.top();str.pop();
13             opa=str.top();str.pop();
14             switch(c[0])  //switch 只能读取一个字符,不能读取字符串,会报错
15            {
16                case '*':
17                {
18                     opa=opa*opb;
19                     str.push(opa);
20                     break;
21                 } 
22                 case '/':
23                 {
24                     opa=opa/opb;
25                     str.push(opa);
26                     break;
27                 }
28                 case '+':
29                 {
30                     opa=opa+opb;
31                     str.push(opa);
32                     break;
33                 } 
34                 case '-':
35                 {
36                     opa=opa-opb;
37                     str.push(opa);
38                     break;
39                 }
40            }
41         }
42            
43         }
44   
45         return str.top();
46     }
47 };
原文地址:https://www.cnblogs.com/daocaorenblog/p/4908856.html