pow(x, n).

Implement pow(x, n).

思路：只用判断n的情况，0,1，-1，然后判断n的奇偶，和符号；

class Solution {
    public double myPow(double x, int n) {
        if(n == 0) return 1.0;
        if(n == 1) return x;
        if(n == -1) return 1.0 / x;
        
        double half = myPow(x, n / 2);
        if(n % 2 == 0) {
            return half * half;
        } else {
            if(n > 0) {
                return x * half * half;
            } else {
                return 1.0 / x * half * half;
            }
        }
    }
}