Divide integers Leetcode

Title: Divide integers

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

If It is overflow, return max_int.

Look at the idea of the great God in the discussion area:

In this problem, we is asked to divide and integers. However, we is not allowed to use division, multiplication and mod operations. So, who else can we use? Yeah, bit manipulations.

Let's do a example and see what bit manipulations work.

Suppose we want to divide 15 3 by, so are and is 15 dividend 3 divisor . Well, division simply requires us to find what many times we can subtract the from the the the divisor dividend without making the c8/> negative.

Let ' s get started. We Subtract3From15And we get12, which is positive. Let's try to subtract more. Well, we shift3To the left by1Bit and we get6. Subtracting6From15Still gives a positive result. Well, we shift again and get12. We Subtract12From15And it is still positive. We shift again, obtaining24And we know we can at the most subtract12. Well, since12is obtained by shifting3To the left twice, we know it is4Times of3. How does we obtain this4? Well, we start from1and shift it to left twice at the same time. We add4To a answer (initialized to be0). In fact, the above process was like15 = 3 * 4 + 3. We now get part of the quotient (4) with a remainder3.

Then we repeat the above process again. We subtract from the divisor = 3 remaining and dividend = 3 obtain 0 . We know we are done. No shift happens, so we simply add to the 1 << 0 answer.

Now we had the full algorithm to perform division.

According to the problem statement, we need to handle some exceptions, such as overflow.

Well, the cases may cause overflow:

1. divisor = 0;
2. dividend = INT_MINand divisor = -1 (because abs(INT_MIN) = INT_MAX + 1 ).

Of course, we also need to take the sign into considerations, which are relatively easy.

Putting all these together, we have the following code.

ClassSolution {Publicint Divide (int dividend,int divisor) {if (!divisor | | (Dividend = = Int_min && divisor = =-1))return int_max; int sign = ((Dividend < 0) ^ (Divisor < 0))?-
                 
                  1: 
                  1; long long DVD = Labs (dividend); long long DVS = Labs (divisor); int res = 0; while (DVD >= DVS) {long long temp = DVS, multiple = 1; while (DVD >= (temp << 1)) {temp <<= 1; Multiple <<= 1;} DVD-= temp; Res + = multiple; } return sign = = 1? Res:-res;}};       
                 

1#include <iostream>2#include <limits>3 using namespacestd;5 classSolution {6  Public:7     intDivideintDividend,intdivisor)8     {9         intSign = ((Dividend >0) ^ (Divisor >0) ? -1:1);Ten         if(!divisor | | (dividend==int_min&&divisor==-1)) One             returnInt_max; A         Long LongdivID = Labs (dividend), Divis =Labs (divisor); -         Long Longres =0; -          while(divID >=Divis) the         { -             Long Longtemp = Divis,multi_time=1; -              while(divID >= (temp<<1)) -             { +Temp <<=1; -Multi_time <<=1; +             } AdivID-=temp; atRes + =Multi_time; -         } -         returnSign = =1? res:-Res; -     } - }; - intMain () in { - solution test; to     intres = Test.divide (0,1); +cout << Res <<Endl; -     return 0; the}

Divide integers Leetcode