河内塔递归java [英] Tower of Hanoi recursion java

问题描述

这是我使用递归解决河内塔的Java代码：

Here's my Java code for solving Tower of Hanoi using recursion:

/**here is a stack of N disks on the first of three poles (call
them A, B and C) and your job is to move the disks from pole A to pole B without
ever putting a larger disk on top of a smaller disk.*/ 

public class Hanoi {

    public static void main(String[] args) {
        playHanoi (2,"A","B","C");
    }

    //move n disks from position "from" to "to" via "other"
    private static void playHanoi(int n, String from , String other, String to) {
        if (n == 0)
            return;
        if (n > 0)
        playHanoi(n-1, from, to, other);
        System.out.printf("Move one disk from pole %s to pole %s \n ", from, to);
        playHanoi(n-1, other, from, to);
    }

}

我放置打印的地方吗方法有关系？另外，我可以这样做：

Does the place I put the print method matter? Also, can I do it like this:

    playHanoi(n-1, from, to, other);
    playHanoi(n-1, other, from, to);
    System.out.printf("Move one disk from pole %s to pole %s \n ", from, to);

推荐答案

解决 Hanoy之塔这样的问题，只不过是定义了你想要完成工作的策略。和你的代码：

Solving Tower of Hanoy problem in this way, is nothing but defining the strategy of how you want to get the job done. And your code :

    playHanoi(n-1, from, to, other);
    System.out.printf("Move one disk from pole %s to pole %s \n ", from, to);
    playHanoi(n-1, other, from, to);

基本上将您的策略​​定义为吼叫，

Basically defines your strategy to like bellow,

1. 从 移动 n-1 磁盘 （源塔） 其他 （中间塔）。


2. 然后移动 n 从 从 （源塔）到 到 （目标塔）的磁盘。


3. 最后从 其他 （中间塔）移动 n-1 磁盘） to （目的地塔）。

1. Move n-1 disks from "from" (source tower) to "other" (intermediary tower).
2. Then move n th disk from "from" (source tower) to "to" (destination tower).
3. Finally move n-1 disks from "other" (intermediary tower) to "to" (destination tower).

您的 prinf 基本上是 2 nd步骤。

Your prinf basically does the 2 nd step.

现在，如果你写这样的代码：

Now if you write code like this:

    playHanoi(n-1, from, to, other);
    playHanoi(n-1, other, from, to);
    System.out.printf("Move one disk from pole %s to pole %s \n ", from, to);

然后你基本上在做：

1. 将 n-1 磁盘从 从 （源塔）移至 其他 （中间塔）。


2. 然后移动 n-1 从 其他 （中间塔）到 到 （目标塔）的磁盘。



3. 最后从 从 移动 n 磁盘（来源）塔）到 到 （目的地塔）。

1. Move n-1 disks from "from" (source tower) to "other" (intermediary tower).
2. Then move n-1 disks from "other" (intermediary tower) to "to" (destination tower).

3. Finally move n th disk from "from" (source tower) to "to" (destination tower).

在此策略中，执行 2 步骤后（从 移动所有 n-1 磁盘）其他 到 到 ）， 3 rd步骤无效（移动 n 磁盘从 从 到 到 ）！因为在 Tower of Hanoy 中你不能把较大的磁盘放在较小的磁盘上！

In this strategy, after doing the 2 nd step (moving all n-1 disks from "other" to "to"), 3 rd step becomes invalid(moving n th disk from "from" to "to")! Because in Tower of Hanoy you cant put a larger disk on a smaller one!

因此，选择第二个选项（策略）会导致您采用无效策略，这就是为什么您不能这样做的原因！

So choosing the second option(strategy) leads you to an invalid strategy, thats why you can not do that!

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