LeetCode OJ：Pow(x, n) （幂运算）

 1 class Solution {
 2 public:
 3     double myPow(double x, int n) {
 4         if(n == INT_MIN)
 5                 return 1.0* x/myPow(x, INT_MAX);
 6         if(n < 0)
 7             return 1.0/myPow(x, -n);
 8         else if(n == 0)
 9             return 1.0;
10         double halfRes = myPow(x, n/2);
11         if(n%2)
12             return halfRes * halfRes * x;
13         return halfRes * halfRes;
14     }
15 };