2021年9月14日21:06:53
递归

public class Solution {
    public double Power(double base, int exponent) {
        if(exponent == 0) return 1;
        else if(exponent>0) return base * Power(base,exponent-1);
        else return 1/base*Power(base,exponent+1);
  }
}

快速幂
图片说明 k神

class Solution {
    public double myPow(double x, int n) {
        if(x == 0) return 0;
        long b = n;
        double res = 1.0;
        if(b < 0) {
            x = 1 / x;
            b = -b;
        }
        while(b > 0) {
            if((b & 1) == 1) res *= x;
            x *= x;
            b >>= 1;
        }
        return res;
    }
}