算法：x的n次方

该题是用来公司教学，并无难度。用于说明算法效率差异以及循环和递归的效率差别。

 

package practice;

import java.math.BigDecimal;

/**
 * @author caiyu
 * @date 2014-12-3
 */
public class X_N_Square {
    static BigDecimal x = new BigDecimal(7);

    public static void main(String[] args) {
        int n = 1000;
        BigDecimal result = x;

        long time = System.currentTimeMillis();
        if (n < 5000)
            for (int i = 1; i < n; i++) {
                result = result.multiply(x);
            }
        System.out.println(System.currentTimeMillis() - time);

        if (n < 5000)
            System.out.println(result);

        time = System.currentTimeMillis();
        result = cal(n, x);
        System.out.println(System.currentTimeMillis() - time);

        if (n < 5000)
            System.out.println(result);

        time = System.currentTimeMillis();
        // 换底公式
        int count = (int) Math.floor(Math.log(n) / Math.log(2));
        result = x;
        while (count-- > 0) {
            result = (n >> count) % 2 == 0 ? result.multiply(result) : result
                    .multiply(result).multiply(x);
        }
        System.out.println(System.currentTimeMillis() - time);

        if (n < 5000)
            System.out.println(result);
    }

    public static BigDecimal cal(int n, BigDecimal r) {
        if (n == 1)
            return x;
        if (n % 2 == 0) {
            r = cal(n / 2, r);
            return r.multiply(r);
        } else {
            r = cal((n - 1) / 2, r);
            return r.multiply(r).multiply(x);
        }
    }

}