剑指offer12：求解double类型的浮点数base和int类型的整数exponent的次方。 保证base和exponent不同时为0

 1 class Solution {
 2 public:
 3     double Power(double base, int exponent) {
 4         int ab_e = std::abs(exponent);
 5 
 6         double res = 1.0;
 7         while(ab_e != 0) {
 8             res *= base;
 9             ab_e--;
10         }
11 
12         if (exponent < 0) {
13             res = 1.0 / res;
14         }
15         return res;
16     }
17 };

 1 class Solution {
 2 public:
 3     double Power(double base, int exponent) {
 4         int ab_e = std::abs(exponent);
 5 
 6         double res = 1.0;
 7         while(ab_e) {
 8             if (ab_e & 1) {
 9                 res *= base;
10             }
11             base *= base;
12             ab_e >>= 1;
13         }
14         if (exponent < 0){
15             res = 1.0 / res;
16         }
17         return res;
18     }
19 };