[LeetCode]Pow(x, n)

 1 class Solution {
 2 public:
 3     double myPow(double x, int n) {
 4         unsigned int exp = (unsigned int)n;
 5         if (n < 0) {
 6             exp = (unsigned int)(-n);
 7             return 1.0 / myPow_l(x, exp);
 8         }
 9         
10         return myPow_l(x, exp);
11     }
12     
13     double myPow_l(double x, unsigned int n) {
14         if (n == 0) {
15             return 1;
16         }
17         
18         if (n == 1) {
19             return x;
20         }
21         
22         double result = myPow(x, n >> 1);
23         result *= result;
24         if ((n & 1) != 0) {
25             result *= x;
26         }
27         
28         return result;
29     }
30 };