LeetCode: pow

Title:

 https://leetcode.com/problems/powx-n/

思路：二分。使用递归或者非递归。非递归有点难理解.pow(0,0)=1 递归的方法是将n为负数的用除法解决。有问题，没有考虑0的负数次幂会导致除数为0.对于非递归，可以这么理解，x是指数，不断增长,x , x^2, x^4,x^8

class Solution {
public:
 
    
    double Pow(double x,int n){

        if (n == 0)
            return 1.0;
        if (n == 1)
            return x;
        double a = Pow(x,n/2);
        if (n % 2 == 0)
            return a*a;
        else
            return a*a*x;
    }
    double myPow(double x,int n){
        if (n < 0){
            if (x == 0.0)
                return 0.0;
            else
                return 1.0/Pow(x,-n);
        }
        else{
            return Pow(x,n);
        }
    }
};

class Solution {
public:
    double myPow(double x, int n) {
        if (0 == n)
            return 1.0;
        if (n < 0){
            if (x == 0)
                return 0.0;
            else{
                x = 1.0 / x;
                n = -1 * n;
            }
        }
        double result = 1.0;
        while (n > 0){
            if (n & 1){
                result *= x;
            }
            n /= 2;
            x *= x;
        }
        return result;
    }
};