数值的整数次方

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

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

 1 class Solution {
 2 public:
 3     double Power(double base, int exponent) {
 4         if(exponent>0)
 5         {
 6             if(exponent==1)
 7                 return base;
 8             if(exponent%2==0)
 9                 return Power(base,exponent/2)*Power(base,exponent/2);
10             else
11                 return Power(base,exponent/2)*Power(base,exponent/2+1);
12         }
13         else if (exponent==0)
14         {
15             return 1;
16         }
17         else
18         {
19             return 1/Power(base,0-exponent);
20         }
21     }
22 };

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 r = 1;
 6         if(base==0) return 0;
 7         while(p){
 8             if(p&1) r *= base;
 9             base *= base;
10             p>>=1;
11         }
12         return exponent<0? 1/r: r;
13     }
14 };