斐波拉契数列不一样的实现

package ch8;

import org.junit.Test;

/**
 * Created by Administrator on 2016/4/11.
 */
public class Main2 {


    /**
     * 传统写法
     * 说明：效率低
     * 时间复杂度：2^n
     */
    public long method1(int n) {
        if (n == 0) {
            return 0;
        }
        if (n == 1) {
            return 1;
        }
        return method1(n - 1) + method1(n - 2);

    }

    /**
     * 加强版写法
     * 时间复杂度：O(n)
     * @param n
     * @return
     */
    public long method2(int n) {
        if(n==0){
            return 0;
        }
        int x = 1;
        int y = 1;
        int tmp;
        for (int i = 2; i < n; i++) {
            tmp = x;
            x = y;
            y = y + tmp;
        }
        return y;
    }

    @Test
    public void test() {
        System.out.println(method1(0));
        System.out.println(method2(0));

    }

}

转载于:https://www.cnblogs.com/ayhq/p/5392784.html