Woodstock Blog

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[CC150v4] 3.4 Towers of Hanoi


In the classic problem of the Towers of Hanoi, you have 3 rods and N disks of di!erent sizes which can slide onto any tower. The puzzle starts with disks sorted in ascending order of size from top to bottom (e.g., each disk sits on top of an even larger one). You have the following constraints:

  1. Only one disk can be moved at a time.
  2. A disk is slid o! the top of one rod onto the next rod.
  3. A disk can only be placed on top of a larger disk.

Write a program to move the disks from the “rst rod to the last using Stacks.


This is a classic recursive question. The solution code is supposed to be very concise.


written by me, slightly different from the answer in the book, but good.

Main class:

public class HanoiMyAnswer {

    private static Rod r0, r1, r2;
    private static final int NUM_DISKS = 5;

    public static void main(String[] args) {
        // Hanoi Tower always have 3 rods
        r0 = new Rod(0);
        r1 = new Rod(1);
        r2 = new Rod(2);

        // Put some disks on the 1st rod, leaving 2nd and 3rd rod empty

        // start main algorithm
        System.out.println("My answer: ");
        moveDisks(NUM_DISKS, r0, r2, r1);

    private static void moveDisks(int number, Rod from, Rod to, Rod buffer) {
        if (number == 1) {
            int topValue = from.disks.pop();
            displayMessage(topValue, from.name, to.name);
        } else {
            moveDisks(number - 1, from, buffer, to);
            int bottomValue = from.disks.pop();
            displayMessage(bottomValue, from.name, to.name);
            moveDisks(number - 1, buffer, to, from);

    private static void displayMessage(int disk, int from, int to) {
        System.out.println("Disk[" + disk + "]: Rod" + from + "-->" + to);


class Rod {

    int name;
    Stack<Integer> disks = new Stack<Integer>();

    public Rod(int rodIndex) {
        this.name = rodIndex;

    public void setDisks(int n) {
        // this method will insert n disks on this Rod
        // the bottom disk is indexed as (n-1) and top one is 0
        for (int i = n - 1; i >= 0; i--) {