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 @ts for adding this problem and creating all test cases.

代码：

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