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 0;
              
          long long result=1;
          unsigned long long tmp=divisor,left;
          
          while(tmp<=dividend){
                 left=dividend-tmp;
                 tmp<<=1;
                 
                 if(tmp>dividend){
                       break;
                 }
                 
                 else{
                       result<<=1;
                 }
          }
          
          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>=0) && (divisor>0)) || ((dividend<=0) && (divisor<0));
          
          return positive?interDivide(_dividend,_divisor):(-1)*interDivide(_dividend,_divisor);
          
    }
};

之所以用unsigned long long是为了防止溢出.