Pow(x, n)

Implement pow(x, n).

public class Solution {
    public double pow(double x, int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int absn = Math.abs(n);
        if (absn==0.0) {
            //0的多少次方都是1
            return 1.0;
        }
        if (absn==1) {
            //一次方是x本身 如果是负数取倒数
            return n>0 ? x:1/x;
        }
        double ret = pow(x,absn/2);
        if (absn%2==0){
            //例如2^4 看做2^2 * 2^2 如果是负数取倒数
            return n>0 ? ret * ret : 1/(ret*ret);
        } else {
            //例如2^5 看做2^2 * 2^2 * 2 如果是负数取倒数
            return n>0 ? ret * ret * x : 1/(ret * ret * x);
        }
    }
}