050. Pow(x, n)

 1 class Solution {
 2 public:
 3     double myPow(double x, int n) {
 4         if (x == 0) return 0;
 5         else {
 6             if (n == 0 || x == 1) return 1;
 7             else {
 8                 if (n > 0) {
 9                     double temp = myPow(x, n / 2);
10                     if (n % 2 == 0) return temp * temp;
11                     else return temp * temp * x;
12                 }
13                 else {
14                     double temp = myPow(x, -(n / 2));
15                     if (n % 2 == 0) return 1.0 / (temp * temp);
16                     else  return 1.0 / (temp * temp * x);
17                 }
18             }
19         }
20     }
21 };

 1 class Solution {
 2 public:
 3     double myPow(double x, int n) {
 4         if (x == 0) return 0;
 5         else {
 6             if (n == 0 || x == 1) return 1;
 7             else {
 8                 if (n > 0) {
 9                     double temp = myPow(x, n / 2);
10                     if (n % 2 == 0) return temp * temp;
11                     else return temp * temp * x;
12                 }
13                 else {
14                     double temp = myPow(x, -(n / 2)); // 这里一定要注意负值的最大可能越界
15                     if (n % 2 == 0) return 1.0 / (temp * temp);
16                     else  return 1.0 / (temp * temp * x);
17                 }
18             }
19         }
20     }
21 };