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📜  河内塔项目

📅  最后修改于: 2021-09-16 10:57:34             🧑  作者: Mango

河内塔是一个数学难题，我们有三个杆和 n 个圆盘。拼图的目标是将整个堆栈移动到另一个杆上，遵循以下简单规则：

  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上面的圆盘并将其放在另一个堆栈的顶部，即圆盘只有在它是堆栈中最上面的圆盘时才能移动。
  3. 任何磁盘都不能放置在较小磁盘的顶部。

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 :

常见问题.disk3

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

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]

视频解说

https://www.youtube.com/watch?v=YstLjLCGmgg

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