LeetCode-Divdend two Integers

题目:

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

思路分析

二分法.将除数不断增倍,而结果同样扩大两倍,直到除数的值大于被除数.然后再利用被除数减去除数最后增长到小于被除数的值，递归求出结果.

例如:123/4

4<123   4*2=8<123  8*2=16>123    16*2=32<123

32*2=64<123   64*2=128>123  故结果增长的值为64/4=16.

再利用123-64=59,再次递归求出结果,最后肯定能得出59/4=14，余数为3,3<4抛弃.

故最终值为16+14=30

代码:

 public Solution{
 public:
     long long  interDivide(unsigned long long dividend,
                       unsigned long long divisor){
           if(dividend<divisor) return ;

           long long result=;
           unsigned long long tmp=divisor,left;

           while(tmp<=dividend){
                  left=dividend-tmp;
                  tmp<<=;

                  if(tmp>dividend){
                        break;
                  }

                  else{
                        result<<=;
                  }
           }

           return result+interDivide(left,divisor);
     }

     int Divide(int dividend,int divisor){
           unsigned long long _dividend=abs((long long)dividend);
           unsigned long long _divisor=abs((long long)divisor);

           bool positive=((dividend>=) && (divisor>)) || ((dividend<=) && (divisor<));

           return positive?interDivide(_dividend,_divisor):(-)*interDivide(_dividend,_divisor);

     }
 };

之所以用unsigned long long是为了防止溢出.