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 <= )
            {
                return n;
            }
            else
            {
                return Foo(n - ) + Foo(n - );
            }
        }

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 = ;
            if (n <= )
            {
                result = n;
            }
            else
            {
                int[] num = new int[n];
                num[] = ;
                num[] = ;

                System.Console.Write(num[] + " " + num[] + " ");
                for (int i = ; i <n-; i++)
                {
                    num[i + ] = num[i + ] + num[i];
                    result = num[i + ];
                    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 <= )
            {
                Console.WriteLine(n);
            }
            else
            {
                int x = , y = ;
                for (int i = ; i <= n; i++, y = x + y, x = y - x)
                {
                    Console.WriteLine(y + " ");
                }
            }
        }