Lintcode-medium-divide integers

Divide-integers without using multiplication, division and mod operator.

If It is overflow, return2147483647

Given dividend = 100 and divisor = 9 , return 11 .

 Public classSolution {/**     * @paramDividend The dividend *@paramdivisor the divisor *@returnThe result*/     Public intDivideintDividend,intdivisor) {        //Write Your code here                if(Divisor = = 0)            return2147483647; if(Dividend = = 0)            return0; if(Divisor = = 1)            returndividend; if(Dividend = = Integer.min_value && Divisor = =-1) {            returnInteger.max_value; }                BooleanNeg = (Dividend < 0 && divisor > 0) | | (Dividend > 0 && Divisor < 0); Longnum = Math.Abs (Long) dividend); LongDen = Math.Abs ((Long) divisor); Longresult = 0;  while(Num >=den) {            intShift = 0;  while(Num >= (den <<shift)) Shift++; Num-= den<< (shift-1); Result+ = 1 << (shift-1); }        if(neg) {if(Result > Math.Abs (Long) integer.min_value))return2147483647; Else                    return(int) (-result); }        Else{            if(Result > (Long) integer.max_value)return2147483647; Else                return(int) result; }    }}

Lintcode-medium-divide integers