[剑指Offer] 12.数值的整数次方

【思路1】递归

 class Solution {
 public:
     double Power(double base, int exponent) {
         if(exponent < ){
             base = /base;
             exponent = -exponent;
         }
         if(exponent == ){
             return ;
         }else{
             return base * Power(base, exponent - );
         }
     }
 };

【思路2】快速幂

 class Solution {
 public:
     double Power(double base, int exponent) {
         long long p = abs((long long)exponent);
         double ans = 1.0;
         while(p != ) {
             if(p & ) {
                 ans *= base;
             }
             base *= base;
             p >>= ;
         }
         return exponent >  ? ans : /ans;
     }
 };