[剑指Offer] 12.数值的整数次方

【思路1】递归

 1 class Solution {
 2 public:
 3     double Power(double base, int exponent) {
 4         if(exponent < 0){
 5             base = 1/base;
 6             exponent = -exponent;
 7         }
 8         if(exponent == 0){
 9             return 1;
10         }else{
11             return base * Power(base, exponent - 1);
12         }
13     }
14 };

【思路2】快速幂

 1 class Solution {
 2 public:
 3     double Power(double base, int exponent) {
 4         long long p = abs((long long)exponent);
 5         double ans = 1.0;
 6         while(p != 0) {
 7             if(p & 1) {
 8                 ans *= base;
 9             }
10             base *= base;
11             p >>= 1;
12         }
13         return exponent > 0 ? ans : 1/ans;
14     }
15 };