Leetcode::pow 

 1 class Solution {
 2 public:
 3     bool equal( double a,double b )
 4     {
 5         if( a-b < 0.0000001 && a-b > -0.0000001 )
 6         {
 7             return true;
 8         }
 9         else
10         {
11             return false;
12         }
13     }
14     
15     double pow_unsigned(double x,unsigned int n )
16     {
17         if( n == 0 )
18         return 1;
19         if( n== 1 )
20         return x;
21         
22         double result=pow_unsigned(x,n>>1);
23         result*=result;
24         if( n & 1 == 1)
25         result*=x;
26         return result;
27     }
28     double pow(double x, int n) {
29         // Start typing your C/C++ solution below
30         // DO NOT write int main() function
31         if( equal(x,0) && n <0 )
32         {
33             return 0.0;
34         }
35         
36         unsigned int t_n=(unsigned int)n;
37         
38         if(n<0)
39         t_n=(unsigned int)(-n);
40         
41         double result=pow_unsigned(x,t_n);
42         
43         if(n<0)
44         result=1.0/result;
45         
46         return result;
47         
48     }
49 };

由于考虑到double、float类型数的精度，equal函数来判断是否等于0，当然直接==判断也行；

时间复杂度为O(logn);

posted @ 2013-07-19 21:38  NinaGood  阅读(116)  评论(0)  编辑  收藏  举报