leetcode_Basic Calculator II

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。