题解 | #数值的整数次方#
数值的整数次方
http://www.nowcoder.com/practice/1a834e5e3e1a4b7ba251417554e07c00
public class Solution {
double a = 1 / base;
int b = exponent * -1;
return a * Power(a,b - 1);
}
if(exponent == 0){
return 1.0;
}
return base * Power(base,exponent - 1);
}
}
public double Power(double base, int exponent) {
//负次幂时
if(exponent < 0){double a = 1 / base;
int b = exponent * -1;
return a * Power(a,b - 1);
}
if(exponent == 0){
return 1.0;
}
return base * Power(base,exponent - 1);
}
}
