Java Program For Selecting A Random Node From A Singly Linked List

Given a singly linked list, select a random node from the linked list (the probability of picking a node should be 1/N if there are N nodes in the list). You are given a random number generator.
Below is a Simple Solution:

1. Count the number of nodes by traversing the list.
2. Traverse the list again and select every node with probability 1/N. The selection can be done by generating a random number from 0 to N-i for i'th node, and selecting the i'th node only if the generated number is equal to 0 (or any other fixed number from 0 to N-i).

We get uniform probabilities with the above schemes.  

i = 1, probability of selecting first node = 1/N i = 2, probability of selecting second node = [probability that first node is not selected] * [probability that second node is selected] = ((N-1)/N)* 1/(N-1) = 1/N 

Similarly, probabilities of other selecting other nodes is 1/N
The above solution requires two traversals of linked list. 

How to select a random node with only one traversal allowed? 
The idea is to use Reservoir Sampling. Following are the steps. This is a simpler version of Reservoir Sampling as we need to select only one key instead of k keys.

(1) Initialize result as first node result = head->key (2) Initialize n = 2 (3) Now one by one consider all nodes from 2nd node onward. (a) Generate a random number from 0 to n-1. Let the generated random number is j. (b) If j is equal to 0 (we could choose other fixed numbers between 0 to n-1), then replace result with the current node. (c) n = n+1 (d) current = current->next

Below is the implementation of above algorithm.

// Java program to select a random  // node from singly linked list import java.util.*; // Linked List Class class LinkedList  {  // head of list  static Node head;   // Node Class   static class Node   {  int data;  Node next;  // Constructor to create   // a new node  Node(int d)   {  data = d;  next = null;  }  }  // A reservoir sampling-based function   // to print a random node from a   // linked list  void printrandom(Node node)   {  // If list is empty  if (node == null)   {  return;  }  // Use a different seed value so   // that we don't get same result   // each time we run this program  Math.abs(UUID.randomUUID().getMostSignificantBits());  // Initialize result as first node  int result = node.data;  // Iterate from the (k+1)th element   // to nth element  Node current = node;  int n;  for (n = 2; current != null; n++)   {  // change result with   // probability 1/n  if (Math.random() % n == 0)   {  result = current.data;  }  // Move to next node  current = current.next;  }  System.out.println(  "Randomly selected key is " + result);  }  // Driver code  public static void main(String[] args)   {  LinkedList list = new LinkedList();  list.head = new Node(5);  list.head.next = new Node(20);  list.head.next.next = new Node(4);  list.head.next.next.next = new Node(3);  list.head.next.next.next.next = new Node(30);  list.printrandom(head);  } } // This code is contributed by Mayank Jaiswal 

Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.

Note that the above program is based on the outcome of a random function and may produce different output.

How does this work? 
Let there be total N nodes in list. It is easier to understand from the last node.
The probability that the last node is result simply 1/N [For last or N'th node, we generate a random number between 0 to N-1 and make the last node as a result if the generated number is 0 (or any other fixed number]
The probability that second last node is result should also be 1/N. 
 

The probability that the second last node is result = [Probability that the second last node replaces result] X [Probability that the last node doesn't replace the result] = [1 / (N-1)] * [(N-1)/N] = 1/N

Similarly, we can show probability for 3^rd last node and other nodes. Please refer complete article on Select a Random Node from a Singly Linked List for more details!