Ouch. I tried to figure out the Towers of Hanoi last year. The tricky thing about TOH is it's not a simple example of recursion - you have nested recursions which also change the roles of towers on each call. The only way I could get it to make sense was to literally visualize the movement of the rings in my mind's eye, and verbalize what the recursive call would be. I would start with a single ring, then two, then three. I actually ordered the game on the internet. It took me maybe two or three days of cracking my brains to get it.

当使用递归解时，我总是尝试:

首先建立基本情况，即。 当n = 1的阶乘 试着想出一个一般的规则 对于其他情况

还有不同类型的递归解，分而治之的方法对分形和其他方法很有用。

如果你能先解决一些简单的问题，只是为了掌握窍门，这也会有帮助。一些例子是求解阶乘和生成第n个斐波那契数。

作为参考，我强烈推荐Robert Sedgewick的算法。

希望这能有所帮助。祝你好运。

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...
    }
}

递归函数只是一个函数，它可以根据需要多次调用自己。如果您需要多次处理某件事，但不确定实际需要多少次，那么它就很有用。在某种程度上，你可以把递归函数看作是一种循环。然而，就像循环一样，您需要指定中断流程的条件，否则它将变得无限。

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。

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.