Common Lisp中的简单递归示例:

MYMAP对列表中的每个元素应用一个函数。

1)空列表没有元素，所以我们返回空列表-()和NIL都是空列表。

2)将函数应用到第一个列表，对列表的其余部分调用MYMAP(递归调用)，并将两个结果合并到一个新列表中。

(DEFUN MYMAP (FUNCTION LIST)
  (IF (NULL LIST)
      ()
      (CONS (FUNCALL FUNCTION (FIRST LIST))
            (MYMAP FUNCTION (REST LIST)))))

让我们观察跟踪执行。在输入函数时，输出参数。在退出函数时，输出结果。对于每个递归调用，输出将按级别缩进。

这个例子对列表(1 2 3 4)中的每个数字调用SIN函数。

Command: (mymap 'sin '(1 2 3 4))

1 Enter MYMAP SIN (1 2 3 4)
| 2 Enter MYMAP SIN (2 3 4)
|   3 Enter MYMAP SIN (3 4)
|   | 4 Enter MYMAP SIN (4)
|   |   5 Enter MYMAP SIN NIL
|   |   5 Exit MYMAP NIL
|   | 4 Exit MYMAP (-0.75680256)
|   3 Exit MYMAP (0.14112002 -0.75680256)
| 2 Exit MYMAP (0.9092975 0.14112002 -0.75680256)
1 Exit MYMAP (0.841471 0.9092975 0.14112002 -0.75680256)

这是我们的结果:

(0.841471 0.9092975 0.14112002 -0.75680256)

你在用哪本书?

关于算法的标准教科书是Cormen & Rivest。我的经验是，它很好地教授了递归。

递归是编程中较难掌握的部分之一，虽然它确实需要本能，但它是可以学习的。但它确实需要一个好的描述，好的例子和好的插图。

此外，30页通常是很多的，30页是用一种编程语言编写的。在你从一本普通的书中理解递归之前，不要尝试用C或Java学习递归。

这与其说是一个问题，不如说是一个抱怨。关于递归你有更具体的问题吗?就像乘法一样，人们不会写很多关于它的东西。

说到乘法，想想这个。

问题:

* b是什么?

答:

如果b = 1，就是a。 否则就是a+a*(b-1)

* (b - 1)是什么?请参考上面的问题来解决这个问题。

构建递归函数的真正数学方法如下:

1:假设你有一个函数对f(n-1)是正确的，构造f使f(n)是正确的。 2:构造f，使得f(1)是正确的。

This is how you can prove that the function is correct, mathematically, and it's called Induction. It is equivalent to have different base cases, or more complicated functions on multiple variables). It is also equivalent to imagine that f(x) is correct for all x Now for a "simple" example. Build a function that can determine if it is possible to have a coin combination of 5 cents and 7 cents to make x cents. For example, it's possible to have 17 cents by 2x5 + 1x7, but impossible to have 16 cents. Now imagine you have a function that tells you if it's possible to create x cents, as long as x < n. Call this function can_create_coins_small. It should be fairly simple to imagine how to make the function for n. Now build your function: bool can_create_coins(int n) { if (n >= 7 && can_create_coins_small(n-7)) return true; else if (n >= 5 && can_create_coins_small(n-5)) return true; else return false; } The trick here is to realize that the fact that can_create_coins works for n, means that you can substitute can_create_coins for can_create_coins_small, giving: bool can_create_coins(int n) { if (n >= 7 && can_create_coins(n-7)) return true; else if (n >= 5 && can_create_coins(n-5)) return true; else return false; } One last thing to do is to have a base case to stop infinite recursion. Note that if you are trying to create 0 cents, then that is possible by having no coins. Adding this condition gives: bool can_create_coins(int n) { if (n == 0) return true; else if (n >= 7 && can_create_coins(n-7)) return true; else if (n >= 5 && can_create_coins(n-5)) return true; else return false; } It can be proven that this function will always return, using a method called infinite descent, but that isn't necessary here. You can imagine that f(n) only calls lower values of n, and will always eventually reach 0. To use this information to solve your Tower of Hanoi problem, I think the trick is to assume you have a function to move n-1 tablets from a to b (for any a/b), trying to move n tables from a to b.

递归

方法A调用方法A调用方法A，最终这些方法A中的一个不会调用并退出，但这是递归，因为有东西调用了它自己。

递归的例子，我想打印出硬盘驱动器上的每个文件夹名称:(在c#中)

public void PrintFolderNames(DirectoryInfo directory)
{
    Console.WriteLine(directory.Name);

    DirectoryInfo[] children = directory.GetDirectories();

    foreach(var child in children)
    {
        PrintFolderNames(child); // See we call ourself here...
    }
}

如果你想要一本很好地用简单的术语解释递归的书，可以看看Gödel，埃舍尔·巴赫:道格拉斯·霍夫施塔特的《永恒的金辫子》，特别是第五章。除了递归，它还能很好地以一种可理解的方式解释计算机科学和数学中的许多复杂概念，一个解释建立在另一个解释的基础上。如果你以前没有接触过这类概念，这可能是一本非常令人兴奋的书。