leetcode[50]Pow(x, n)

Implement pow(xn).

class Solution {
public:
    double pow(double x, int n) {
        if(n==0)return 1.0;
        if(n<0)
        {
            double half=pow(x,-n/2);
            if(-n%2)return 1/(half*half*x);
            else return 1/(half*half);
        }
        return 1.0/pow(x,-n);
        if(n>0)
        {
            double half=pow(x,n/2);
            if(n%2)return half*half*x;
            else return half*half;
        }
    }
};
原文地址:https://www.cnblogs.com/Vae1990Silence/p/4283550.html