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:

 1 public Solution{
 2 public:
 3     long long  interDivide(unsigned long long dividend,
 4                       unsigned long long divisor){
 5           if(dividend<divisor) return 0;
 6               
 7           long long result=1;
 8           unsigned long long tmp=divisor,left;
 9           
10           while(tmp<=dividend){
11                  left=dividend-tmp;
12                  tmp<<=1;
13                  
14                  if(tmp>dividend){
15                        break;
16                  }
17                  
18                  else{
19                        result<<=1;
20                  }
21           }
22           
23           return result+interDivide(left,divisor);
24     }
25     
26     int Divide(int dividend,int divisor){
27           unsigned long long _dividend=abs((long long)dividend);
28           unsigned long long _divisor=abs((long long)divisor);
29           
30           bool positive=((dividend>=0) && (divisor>0)) || ((dividend<=0) && (divisor<0));
31           
32           return positive?interDivide(_dividend,_divisor):(-1)*interDivide(_dividend,_divisor);
33           
34     }
35 };

之所以用unsigned long long是为了防止溢出.