河内塔计划

河内之塔是一个数学难题，其中我们有三个杆和n个盘。难题的目的是遵循以下简单规则将整个堆栈移动到另一根杆：

1. 一次只能移动一个磁盘。
2. 每次移动都包括从一个堆栈中取出较高的磁盘并将其放在另一个堆栈的顶部，即，只有在磁盘是堆栈中最上面的磁盘时，才能移动该磁盘。
3. 请勿在较小的磁盘上放置任何磁盘。

方法 ：

Take an example for 2 disks :
Let rod 1 = 'A', rod 2 = 'B', rod 3 = 'C'.

Step 1 : Shift first disk from 'A' to 'B'.
Step 2 : Shift second disk from 'A' to 'C'.
Step 3 : Shift first disk from 'B' to 'C'.

The pattern here is :
Shift 'n-1' disks from 'A' to 'B'.
Shift last disk from 'A' to 'C'.
Shift 'n-1' disks from 'B' to 'C'.

Image illustration for 3 disks :

例子：

Input : 2
Output : Disk 1 moved from A to B
         Disk 2 moved from A to C
         Disk 1 moved from B to C

Input : 3
Output : Disk 1 moved from A to C
         Disk 2 moved from A to B
         Disk 1 moved from C to B
         Disk 3 moved from A to C
         Disk 1 moved from B to A
         Disk 2 moved from B to C
         Disk 1 moved from A to C

C++

// C++ recursive function to
// solve tower of hanoi puzzle
#include 
using namespace std;
 
void towerOfHanoi(int n, char from_rod,
                    char to_rod, char aux_rod)
{
    if (n == 1)
    {
        cout << "Move disk 1 from rod " << from_rod <<
                            " to rod " << to_rod<

---

Java

// JAVA recursive function to
// solve tower of hanoi puzzle
import java.util.*;
import java.io.*;
import java.math.*;
class GFG
{
static void towerOfHanoi(int n, char from_rod,
                    char to_rod, char aux_rod)
{
    if (n == 1)
    {
        System.out.println("Move disk 1 from rod "+
                           from_rod+" to rod "+to_rod);
        return;
    }
    towerOfHanoi(n - 1, from_rod, aux_rod, to_rod);
    System.out.println("Move disk "+ n + " from rod " +
                       from_rod +" to rod " + to_rod );
    towerOfHanoi(n - 1, aux_rod, to_rod, from_rod);
}
 
// Driver code
public static void  main(String args[])
{
    int n = 4; // Number of disks
    towerOfHanoi(n, 'A', 'C', 'B'); // A, B and C are names of rods
}
}
 
// This code is contributed by jyoti369

Python3

# Recursive Python function to solve tower of hanoi
 
def TowerOfHanoi(n , from_rod, to_rod, aux_rod):
    if n == 1:
        print("Move disk 1 from rod",from_rod,"to rod",to_rod)
        return
    TowerOfHanoi(n-1, from_rod, aux_rod, to_rod)
    print("Move disk",n,"from rod",from_rod,"to rod",to_rod)
    TowerOfHanoi(n-1, aux_rod, to_rod, from_rod)
         
# Driver code
n = 4
TowerOfHanoi(n, 'A', 'C', 'B')
# A, C, B are the name of rods
 
# Contributed By Harshit Agrawal

C#

// C# recursive program to solve tower of hanoi puzzle
using System;
class GFG
{
    static void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod)
    {
        if (n == 1)
        {
            Console.WriteLine("Move disk 1 from rod " + 
                              from_rod + " to rod " + to_rod);
            return;
        }
        towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
        Console.WriteLine("Move disk " + n + " from rod " + 
                          from_rod + " to rod " + to_rod);
        towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
    }
     
    //  Driver method
    public static void Main(String []args)
    {
        int n = 4; // Number of disks
        towerOfHanoi(n, 'A', 'C', 'B');  // A, B and C are names of rods
    }
}
 
//This code is contributed by shivanisinghss2110

PHP

 0){
        TOH($n-1, $A, $C, $B);
        echo "Move disk from rod $A to rod $C \n";
        move($A, $C);
        dispPoles();
        TOH($n-1, $B, $A, $C);
    }
    else {
        return;
    }
}
 
function initPoles($n){
    global $poles;
 
    for ($i=$n; $i>=1; --$i){
        $poles[0][] = $i;
    }
}
 
 
function move($source, $destination){
    global $poles;
     
    // get source and destination pointers
    if ($source=="A") $ptr1=0;
    elseif ($source=="B") $ptr1 = 1;
    else $ptr1 = 2;
     
    if ($destination=="A") $ptr2 = 0;
    elseif ($destination=="B") $ptr2 = 1;
    else $ptr2 = 2;
     
    $top = array_pop($poles[$ptr1]);
    array_push($poles[$ptr2], $top);
}
 
function dispPoles(){ 
    global $poles;
    echo "A: [".implode(", ", $poles[0])."] ";
    echo "B: [".implode(", ", $poles[1])."] ";
    echo "C: [".implode(", ", $poles[2])."] ";
    echo "\n\n";
}
 
$numdisks = 4;
initPoles($numdisks);
echo "Tower of Hanoi Solution for $numdisks disks: \n\n";
dispPoles();
TOH($numdisks);
 
// This code is contributed by ShreyakChakraborty
?>

Javascript

// javascript recursive function to
// solve tower of hanoi puzzle
function towerOfHanoi(n, from_rod,  to_rod,  aux_rod)
{
        if (n == 1)
        {
            document.write("Move disk 1 from rod " + from_rod +
            " to rod " + to_rod+"<br/>");
            return;
        }
        towerOfHanoi(n - 1, from_rod, aux_rod, to_rod);
        document.write("Move disk " + n + " from rod " + from_rod +
        " to rod " + to_rod+"<br/>");
        towerOfHanoi(n - 1, aux_rod, to_rod, from_rod);
    }
 
    // Driver code
    var n = 4; // Number of disks
    towerOfHanoi(n, 'A', 'C', 'B'); // A, B and C are names of rods
 
// This code is contributed by gauravrajput1

输出

Move disk 1 from rod A to rod B
Move disk 2 from rod A to rod C
Move disk 1 from rod B to rod C
Move disk 3 from rod A to rod B
Move disk 1 from rod C to rod A
Move disk 2 from rod C to rod B
Move disk 1 from rod A to rod B
Move disk 4 from rod A to rod C
Move disk 1 from rod B to rod C
Move disk 2 from rod B to rod A
Move disk 1 from rod C to rod A
Move disk 3 from rod B to rod C
Move disk 1 from rod A to rod B
Move disk 2 from rod A to rod C
Move disk 1 from rod B to rod C

Output:

Tower of Hanoi Solution for 4 disks:

A: [4, 3, 2, 1] B: [] C: []

Move disk from rod A to rod B
A: [4, 3, 2] B: [1] C: []

Move disk from rod A to rod C
A: [4, 3] B: [1] C: [2]

Move disk from rod B to rod C
A: [4, 3] B: [] C: [2, 1]

Move disk from rod A to rod B
A: [4] B: [3] C: [2, 1]

Move disk from rod C to rod A
A: [4, 1] B: [3] C: [2]

Move disk from rod C to rod B
A: [4, 1] B: [3, 2] C: []

Move disk from rod A to rod B
A: [4] B: [3, 2, 1] C: []

Move disk from rod A to rod C
A: [] B: [3, 2, 1] C: [4]

Move disk from rod B to rod C
A: [] B: [3, 2] C: [4, 1]

Move disk from rod B to rod A
A: [2] B: [3] C: [4, 1]

Move disk from rod C to rod A
A: [2, 1] B: [3] C: [4]

Move disk from rod B to rod C
A: [2, 1] B: [] C: [4, 3]

Move disk from rod A to rod B
A: [2] B: [1] C: [4, 3]

Move disk from rod A to rod C
A: [] B: [1] C: [4, 3, 2]

Move disk from rod B to rod C
A: [] B: [] C: [4, 3, 2, 1]

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