[LeetCode] 50. Pow(x, n) Java

题目：

Implement pow(x, n).

Example 1:

Input: 2.00000, 10
Output: 1024.00000

Example 2:

Input: 2.10000, 3
Output: 9.26100

题意及分析：实现求x的n次方，使用分治法，复杂度降低到log2n

代码：

public class Solution {
    public double myPow(double x, int n) {
        if(n < 0)
            return 1/pow(x,-n);
        else
            return pow(x,n);
    }

    private double pow(double x, int n) {        //计算x的n次方
        if(n==0) return 1.0;
        double tmp=pow(x,n/2);
        if(n%2==0)
            return tmp*tmp;
        else
            return tmp*tmp*x;
    }
}