Add function documentation, enhance comments in TowerOfHanoi.java (#5533)

This commit is contained in:
Hardik Pawar
2024-10-13 15:00:53 +05:30
committed by GitHub
parent 6af7f7b126
commit 596c6147af
3 changed files with 116 additions and 19 deletions

View File

@ -1,29 +1,65 @@
package com.thealgorithms.others;
import java.util.Scanner;
import java.util.List;
/**
* The {@code TowerOfHanoi} class provides a recursive solution to the Tower of Hanoi puzzle.
* This puzzle involves moving a set of discs from one pole to another, following specific rules:
* 1. Only one disc can be moved at a time.
* 2. A disc can only be placed on top of a larger disc.
* 3. All discs must start on one pole and end on another.
*
* This implementation recursively calculates the steps required to solve the puzzle and stores them
* in a provided list.
*
* <p>
* For more information about the Tower of Hanoi, see
* <a href="https://en.wikipedia.org/wiki/Tower_of_Hanoi">Tower of Hanoi on Wikipedia</a>.
* </p>
*
* The {@code shift} method takes the number of discs and the names of the poles,
* and appends the steps required to solve the puzzle to the provided list.
* Time Complexity: O(2^n) - Exponential time complexity due to the recursive nature of the problem.
* Space Complexity: O(n) - Linear space complexity due to the recursion stack.
* Wikipedia: https://en.wikipedia.org/wiki/Tower_of_Hanoi
*/
final class TowerOfHanoi {
private TowerOfHanoi() {
}
public static void shift(int n, String startPole, String intermediatePole, String endPole) {
// if n becomes zero the program returns thus ending the loop.
/**
* Recursively solve the Tower of Hanoi puzzle by moving discs between poles.
*
* @param n The number of discs to move.
* @param startPole The name of the start pole from which discs are moved.
* @param intermediatePole The name of the intermediate pole used as a temporary holding area.
* @param endPole The name of the end pole to which discs are moved.
* @param result A list to store the steps required to solve the puzzle.
*
* <p>
* This method is called recursively to move n-1 discs
* to the intermediate pole,
* then moves the nth disc to the end pole, and finally
* moves the n-1 discs from the
* intermediate pole to the end pole.
* </p>
*
* <p>
* Time Complexity: O(2^n) - Exponential time complexity due to the recursive nature of the problem.
* Space Complexity: O(n) - Linear space complexity due to the recursion stack.
* </p>
*/
public static void shift(int n, String startPole, String intermediatePole, String endPole, List<String> result) {
if (n != 0) {
// Shift function is called in recursion for swapping the n-1 disc from the startPole to
// the intermediatePole
shift(n - 1, startPole, endPole, intermediatePole);
System.out.format("Move %d from %s to %s%n", n, startPole, endPole); // Result Printing
// Shift function is called in recursion for swapping the n-1 disc from the
// intermediatePole to the endPole
shift(n - 1, intermediatePole, startPole, endPole);
// Move n-1 discs from startPole to intermediatePole
shift(n - 1, startPole, endPole, intermediatePole, result);
// Add the move of the nth disc from startPole to endPole
result.add(String.format("Move %d from %s to %s", n, startPole, endPole));
// Move the n-1 discs from intermediatePole to endPole
shift(n - 1, intermediatePole, startPole, endPole, result);
}
}
public static void main(String[] args) {
System.out.print("Enter number of discs on Pole 1: ");
Scanner scanner = new Scanner(System.in);
int numberOfDiscs = scanner.nextInt(); // input of number of discs on pole 1
shift(numberOfDiscs, "Pole1", "Pole2", "Pole3"); // Shift function called
scanner.close();
}
}