Lintcode Medium title: Divide Two integers two number division

Topic

divide two integers

Dividing two integers requires the multiplication, division, and MOD operators not to be used.

If overflow, returns 2147483647 .

Sample Example

Given divisor = 100 , divisor = 9 , return11

Solving

15% pass rate, subtraction, bitwise operation? I don't know how to do that.

Law one: The use of subtraction, time-out, manual removal of some of the situation is too rogue.

 Public classSolution {/**     * @paramDividend The dividend *@paramdivisor the divisor *@returnThe result*/     Public intDivideintDividend,intdivisor) {        //Write Your code here        if(Dividend = = 0)            return0; if(Dividend = =divisor)return1; intCount = 0; intFlag1 = 1; intFlag2 = 1; if(Dividend = = Integer.min_value && divisor ==1)                returndividend; if(Dividend = = Integer.min_value && Divisor = =-1)                return2147483647; if(dividend<0) {Flag1=-1; Dividend= -dividend; }        if(divisor<0) {Flag2=-1; Divisor= -Divisor; }        //dividend = 2147483647; //divisor = 2;        if(Divisor = = 1)            returnDividend*flag1*Flag2;  while(Dividend >=divisor) {Dividend-=Divisor; Count+=1; }        returnCount*flag1*Flag2; }}

Java Code

Law II: Batch subtraction, reference blog, but also the supermarket, the divisor is 1 time out, is 1 when I deal with alone, is 2 time out

 Public classSolution {/**     * @paramDividend The dividend *@paramdivisor the divisor *@returnThe result*/     Public intDivideintDividend,intdivisor) {        //Write Your code here        if(Dividend = = 0)            return0; if(Dividend = =divisor)return1; intCount = 0; intFlag1 = 1; intFlag2 = 1; if(Dividend = = Integer.min_value && divisor ==1)                returndividend; if(Dividend = = Integer.min_value && Divisor = =-1)                return2147483647; if(dividend<0) {Flag1=-1; Dividend= -dividend; }        if(divisor<0) {Flag2=-1; Divisor= -Divisor; }        //dividend = 2147483647; //divisor = 2;        if(Divisor = = 1)            returnDividend*flag1*Flag2;  while(Dividend >=divisor) {            intsum =Divisor; intCount1 = 1;  while(Sum + sum <=dividend) {Count1+=Count1; Sum+=sum; } Dividend-=sum; Count+=Count1; }        returnCount*flag1*Flag2; }}

Java Code

Method three: Using bit arithmetic

Reference blog, feel and law two very much like, I put two in the weight loss change bit operation, also run timeout, I do not understand, directly with his program can pass.

 Public classSolution {/**     * @paramDividend The dividend *@paramdivisor the divisor *@returnThe result*/     Public intDivideintDividend,intdivisor) {        //Write Your code here        if(Divisor ==0)            returnInteger.max_value; if(Divisor = =-1 && Dividend = =integer.min_value)returnInteger.max_value; intCount = 0; LongPdividend = Math.Abs ((Long) dividend); LongPdivisor = Math.Abs ((Long) divisor);  while(Pdividend >=pdivisor) {            intCount1 = 0;  while((PDIVISOR&LT;&LT;COUNT1) <=pdividend) {Count1++; } Count+ = 1<< (count1-1); Pdividend-= (pdivisor<< (count1-1)); }        if(Dividend >0 && divisor >0 | | dividend<0 && divisor<0)            returncount; Else            return-count; }}

Java Code

The bit operations in the blog can also be changed to subtraction, that is, I do the subtraction is to pass, can not understand

 Public classSolution {/**     * @paramDividend The dividend *@paramdivisor the divisor *@returnThe result*/     Public intDivideintDividend,intdivisor) {        //Write Your code here        if(Divisor ==0)            returnInteger.max_value; if(Divisor = =-1 && Dividend = =integer.min_value)returnInteger.max_value; intCount = 0; LongPdividend = Math.Abs ((Long) dividend); LongPdivisor = Math.Abs ((Long) divisor);  while(Pdividend >=pdivisor) {            intCount1 = 1; Longsum =Pdivisor;  while(sum + sum) <=pdividend) {Count1+=Count1; Sum+=sum; } Count+=Count1; Pdividend-=sum; }        if(Dividend >0 && divisor >0 | | dividend<0 && divisor<0)            returncount; Else            return-count; }}

Java Code

Lintcode Medium title: Divide Two integers two number division