"Leetcode 29" Divide, integers

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

If It is overflow, return max_int.

Test instructions

Achieve division, but do not allow multiplication, division, and modulo operations.

Ideas:

A bit less must appear very low, in fact can be used for the first time dividend minus divisor, the second time with Dividend-divisor minus 2 * divisor, the third time with dividend-3 * divisor minus 4 * divisor ... Until the reduction, then at this time obediently with dividend-(x = 1+2+4+...+2^n) * Divisor minus divisor,dividend-(x+1) * Divisor minus 2 * divisor ... Cycle, count the good number is OK. This problem has a special trick, online a lot of accept code I submit already had to pass (very curious how they did ...) ), can only be manually judged in silence.

C++:

1 classSolution {2  Public:3     intDivideintDividend,intdivisor) {4         5         if(Dividend = =0|| divisor = =0)6             return 0;7         //special judgment, exceeding maximum value should return Max_int8         if(Dividend = =-2147483648&& divisor = =-1)9             return 2147483647;Ten          One         Long LongA = ABS (static_cast<Long Long>(dividend)); A         Long Longb = ABS (static_cast<Long Long>(divisor)); -         Long LongRET =0; -          the          while(A >=b) -         { -             Long Longc =b; -              +              for(inti =0; a >= C; i++) -             { +A-=C; AC <<=1; atRET + =1<<i; -             } -         } -         return((dividend ^ divisor) >> to) ? (-ret): (ret); -     } -};

"Leetcode 29" Divide, integers