[Leetcode] Divide integers

Divide integers divide, integers without using multiplication, Division and mod Operator. if it is OVERFL ow, return max_int.  idea: similar to binary search algorithm, but we do is to multiply or subtract the divisor. Note: 1, in advance to determine the divisor and the divisor of positive and negative, used to determine the positive and negative results of the final result. 2, in order to prevent overflow, the int type is converted to a long long type for operation, and the result is converted to int type. For example:100/3  the first step: because of the > 3, get dividend = 100-3 = 6, Dividor = 3 + 3 = 1, Pre_ret = 1, ret =; the second step: because of the > 6, To dividend = 97-6 = Dividend, Dividor = 6 + 6 = N, Pre_ret = 1 + 1 = 2, ret = 1 + 2 = 3; step three: Due to the > 12, the Ividor = 24, Pre_ret = 4, ret = 7; Fourth step: Because the >, get dividend =, Dividor =, Pre_ret = 8, ret = 15; Fifth step: Because of > 48 , Get dividend = 7, Dividor = 7, Pre_ret = +, ret = 31;  (because at this time dividend > Dividor, so the multiplication of dividor here ends, the next do the double minus) sixth step: Because < 96, Get dividend = 7, Dividor =, Pre_ret = 8, ret = 31; Fourth Step: Because 7 < 48, get dividend = 7, Dividor =, Pre_ret = 4, ret = 31; Fourth Step: Because 7 < 24, get dividend = 7, Dividor =, Pre_ret = 2, ret = 31, Fourth step: Because 7 < 12, get dividend = 7, Dividor = 6, Pre_ret = 1, ret = 31; Fourth step: Because 7 > 6, get dividend = 1, Dividor = 3, Pre_ret = 0, ret = 33; At this time 1 is less than the most primitive divisor 3, terminating the loop, and returning RET according to the positive and negative relationship of divisor and divisor Values.  

1 classSolution2 {3  Public:4   intDivideintDividend,intDividor)5   {6     if(Dividend = = Int_min && Dividor = =-1)7       returnInt_max;8  9Auto sign = [=] (intx)int{returnX>0?1: -1;};Ten     intP1 =Sign (dividend); One     intP2 =Sign (Dividor); A   -     Long LongN1 = ABS (static_cast<Long Long>(dividend)); -     Long LongN2 = ABS (static_cast<Long Long>(Dividor)); the   -     Long LongRET =0; -     Long LongPre_ret =0; -      while(N1 >=n2) +     { -       if(ret = =0&& Pre_ret = =0) +       { APre_ret =1; atRET =1; -       } -       Else -       { -Pre_ret + =Pre_ret; -RET = Pre_ret +ret; in       } -   toN1-=N2; +N2 + =N2; -     } the   *      while(N2 >= Abs (static_cast<Long Long> (Dividor)) && N1 >= ABS (static_cast<Long Long>(Dividor))) $     {Panax Notoginseng       if(Pre_ret = =0) -       { theRET + =1; +          Break; A       } the   +       if(N1 <n2) -       { $N2 = N2 >>1; $         if(N1 = =n2) -         { -RET + =Pre_ret; the            Break; -         }WuyiPre_ret = Pre_ret >>1; the       } -       Else Wu       { -N1-=N2; AboutN2 = N2 >>1; $         if(N1 = =n2) -         { -RET + =Pre_ret; -            Break; A         } +ret = Pre_ret + Pre_ret +ret; thePre_ret = Pre_ret >>1; -       } $     } the   the     if(P1 = =p2) the       returnret; the     Else -       return-ret; in   } the};

[Leetcode] Divide integers