Java思路:
直接上快速幂
class Solution {
public static double ksm(double a, int b) {
double ans = 1;
while(b != 0) {
if((b & 1) != 0) ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}
public double Power(double base, int exponent) {
return exponent > 0 ? ksm(base, exponent) : 1 / ksm(base, -exponent);
}
}
妙啊,比其他人递归的简单易懂