题目：

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7" 3/2 " = 1" 3+5 / 2 " = 5

Note: Do not use the eval built-in library function.

Credits:

Special thanks to  for adding this problem and creating all test cases.

代码：

class Solution {public:    int calculate(string s) {    stack
    
      st;    int result = 0;    int i = 0;    while(i
     
      ='0'&&s.at(i)<='9')         {             while(i
      
       ='0'&&s.at(i)<='9')             {               sum = sum * 10 + s.at(i)-'0';               i++;               }             st.push(sum);         }         else if(i
       
        ='0'&&s.at(i)<='9')                  {                      while(i
        
         ='0'&&s.at(i)<='9')                      {                         sum1 = sum1 * 10 + s.at(i)-'0';                         i++;                                }                      a*=sum1;                      st.push(a);                    }         }         else if(i
         
          ='0'&&s.at(i)<='9') { sum1 = sum1 * 10 + s.at(i)-'0'; i++; } a = (int)(a/sum1); st.push(a); } else { i++; } } stack
          
            st1; while(!st.empty()) { st1.push(st.top()); st.pop(); } while(!st1.empty()) { int a = st1.top(); st1.pop(); if(st1.empty()) return a; else { if((char)(st1.top())=='+') { st1.pop(); int temp = st1.top()+a; st1.pop(); st1.push(temp); } else if((char)(st1.top())=='-') { st1.pop(); int temp = a-st1.top(); st1.pop(); st1.push(temp); } } } return st1.top(); }};
          
         
        
       
      
     
    

方法比较繁琐。。。就这样了。