/**

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

 *

 * Created by lverpeng on 2017/7/18.

 *

 * Implement pow(x, n).

 *

 */

public class Pow {

    /**

     * 实现x^n

     * 考虑n为负数

     * 当n为偶数的时候，计算x*x，以期让n快速收缩

     *

     * @param x

     * @param n

     * @return

     */

    public double pow (int x, int n) {

        boolean sign = true;        // true：正数

        if (n < 0) {

            n = -n;

            sign = false;

        }

        double result = 1.0;

        while (n > 0) {

            if ((n & 1) == 1) {

                result *= x;

            }

            n = n >> 1;

            x *= x;

        }

        return sign ? result : 1.0 / result;

    }

    public static void main(String[] args) {

        Pow pow = new Pow();

        System.out.println(pow.pow(2, 3));

        System.out.println(pow.pow(2, 10));

        System.out.println(pow.pow(2, -3));

    }

}