Implement pow(x, n).

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

思路：不要连续乘n次，会超时。递归求解，注意不要写成重复计算，有正负数处理。

public class Solution {	public double pow(double x, int n) {		if (n == 0)			return 1;		double res = 1;		long limit = n;		limit = limit < 0 ? -limit : limit;		res = recursivePow(x, limit);		if (n < 0)			res = 1.0 / res;		return res;	}	private double recursivePow(double x, long n) {		if (n == 1)			return x;		double half = recursivePow(x, n / 2);		if (n % 2 == 0) {			return half * half;		} else {			return half * half * x;		}	}	public static void main(String[] args) {		System.out.println(new Solution().pow(1.00000, -2147483648));	}}