"Leetcode" 227. Basic Calculator

Problem:

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.

Assume that the given expression was always valid.

Some Examples:

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

Note: Don't use the built-in library function. eval

Solution:

The string is divided into two cases: the number num and the operator op, the result is set to Res, taking into account the algorithm of first multiplication after adding and subtraction, set the middle number of mid-first calculate multiplication , and then add to Res .

*> C + + version

1 classSolution {2  Public:3     intCalculatestrings)4     {5        intres=0, num=0, op='+', i=0, mid=0;6         while(i<s.size ())7        {8num =0;////each round to initialize NUM9            //get an entire number stored in NumTen            if(IsDigit (S[i])) One            { A                  while(I<s.size () &&isdigit (S[i]))//consecutive numeric characters for a number -                 {    -num = num*Ten+ (s[i]-'0'); thei++; -                 } -          -                //The get operator is stored in the OP +                 if(op=='+'|| op=='-') -                 { +res = res+mid; AMid = num* (op=='-'?-1:1); at                 }      -                 Else if(op=='/') -                 { -MID = Mid/num; -                 } -                 Else if(op=='*') in                 { -MID = mid*num; to                 } +            } -            Else if(s[i]==' ')//do not miss the opening space this case =-= It meow was put together thei++; *            Else      $            {Panax Notoginsengop=S[i]; -i++; the            } +        }  Ares = res+mid; the         returnRes; +     } -};

"Leetcode" 227. Basic Calculator