【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 };