11 数值的整数次方

给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。

比如求2的32次方，先求出2的16次方，然后2^16 * 2^16
然后递归求2^16，如果次方为奇数，则再乘一个base

C++：

 1 class Solution {
 2 private:
 3     bool invalidInput = false ;
 4 public:
 5     bool Equal(double num1 , double num2){
 6         if ((num1-num2)<0.0000001 && (num1-num2)>-0.0000001)
 7             return true ;
 8         else
 9             return false ;
10     }
11 
12     double PowerWithUnsignedExponent(double base , int exponent){
13         if (exponent == 0)
14             return 1 ;
15         if (exponent == 1)
16             return base ;
17         double result = PowerWithUnsignedExponent(base,exponent>>1) ;
18         result *=result ;
19         if (exponent&1)
20             result *= base ;
21         return result ;
22     }
23 
24     double Power(double base , int exponent){
25         bool invalidInput = false ;
26         if (Equal(base,0.0) && exponent < 0){
27             invalidInput = true ;
28             return 0.0 ;
29         }
30         int absExponent = abs(exponent) ;
31         double result = PowerWithUnsignedExponent(base,absExponent) ;
32         if (exponent < 0)
33             result = 1.0/result ;
34         return result ;
35     }
36 };

java:

 1 public class Solution {
 2     public double Power(double base, int exponent) {
 3         if (exponent == 0)
 4             return 1 ;
 5         if (exponent == 1)
 6             return base ;
 7         boolean flag = false ;
 8         if (exponent < 0){
 9             exponent = -exponent ;
10             flag = true ;
11         }
12         double result = Power(base * base , exponent/2) ;
13         if (exponent%2 != 0){
14             result *= base ;
15         }
16         if (flag){
17             result = 1/result ;
18         }
19         return result ;
20   }
21 }