*Pow(x, n)

Implement pow(x, n).

public class Solution {
        public double myPow(double x, int n) {
        if(n == 0)
            return 1;
        if(n<0){
            n = -n;
            x = 1/x;
        }
        return (n%2 == 0) ? myPow(x*x, n/2) : x*myPow(x*x, n/2);
    }
}

https://leetcode.com/discuss/17005/short-and-easy-to-understand-solution

【推广】 免费学中医，健康全家人
原文地址：https://www.cnblogs.com/hygeia/p/5090544.html