求1,1,2,3,5,8,13 斐波那契数列第N个数的值

 朋友问了个斐波那契算法。我给出了个递归算法

public static int Foo(int n)
        {
            if (n <= 2)
            {
                return n;
            }           
            else
            {
                return Foo(n - 1) + Foo(n - 2);
            }
        }

结果被打击了，说递归效率不行啊，于是网上饿补了下，发现很多方法，总结了两个

1. 递推工式 num[i + 2] = num[i + 1] + num[i];

        public static int GetNum1(int n)
        {
            int result = 0;
            if (n <= 2)
            {
                result = n;
            }
            else
            {
                int[] num = new int[n];
                num[0] = 1;
                num[1] = 1;

                System.Console.Write(num[0] + " " + num[1] + " ");
                for (int i = 0; i <n-2; i++)
                {
                    num[i + 2] = num[i + 1] + num[i];
                    result = num[i + 2];
                    System.Console.Write(result + "   ");
                }
            }

            return result;
        }

2. 更简单的算法，这个很高级哦

y就是当前第n个值

x就是当前第n-1个的值

        public static void GetNum2(int n)
        {
            if (n <= 2)
            {
                Console.WriteLine(n);
            }
            else
            {
                int x = 0, y = 1;
                for (int i = 1; i <= n; i++, y = x + y, x = y - x)
                {
                    Console.WriteLine(y + " ");
                }
            }
        }