DivideTwoIntegers

//Divide Two Integers Total Accepted: 34192 Total Submissions: 223949 My Submissions Question Solution 
//Divide two integers without using multiplication, division and mod operator.
//
//If it is overflow, return MAX_INT.

下面时申请动态数组来存 divisor*2^i的值，以便来与dividend进行比较来确定要不要这个值，适用与结果值比较小的情况，而直接用 divisor*2^31来做不断递减，适用于值大的情况。
public class DivideTwoIntegers {


public static void main(String[] args) {
// TODO 自动生成的方法存根
System.out.print(divide(-2147483648, -1));
// long n=2147483648L;
// System.out.print((int)n);


}
public static int divide(int dividend, int divisor) {
if(divisor==0||dividend==0)
return 0;

       boolean positive=true;
       if((dividend>0&&divisor<0)||(dividend<0&&divisor>0))
        positive=false;
       long did=dividend>0?(long)dividend:-(long)dividend;
       long dir=divisor>0?(long)(divisor):-(long)divisor;
       long quotients=positiveDivide(did,dir);
       //-2147483648, -1
       if(positive&"ients>Integer.MAX_VALUE)
        return Integer.MAX_VALUE;
       return positive?(int)quotients:-(int)quotients;
}
private static long positiveDivide(long did, long dir) {
// TODO 自动生成的方法存根
long []array=new long[32];
long sum=0;
int i=1;
long quotients=0;
//-2147483648, 1
if(dir==1)
return did;
for(array[0]=dir;array[i-1]<=did;i++)
array[i]=array[i-1]<<1;
for(i=i-2;i>=0;i--){
if(sum<=did-array[i]){
sum+=array[i];
quotients+=1< }
}
return quotients;
}


}