package com.freetymekiyan.algorithms.level.medium; import com.freetymekiyan.algorithms.utils.Utils.ListNode; import java.util.Random; /** * Given a singly linked list, return a random node's value from the linked list. Each node must have the same * probability of being chosen. * <p> * Follow up: * What if the linked list is extremely large and its length is unknown to you? Could you solve this efficiently without * using extra space? * <p> * Example: * <p> * // Init a singly linked list [1,2,3]. * ListNode head = new ListNode(1); * head.next = new ListNode(2); * head.next.next = new ListNode(3); * Solution solution = new Solution(head); * <p> * // getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning. * solution.getRandom(); * <p> * Tags: Reservoir Sampling */ public class LinkedListRandomNode { /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { private final ListNode head; private final Random r; private ListNode current; /** * @param head The linked list's head. * Note that the head is guaranteed to be not null, so it contains at least one node. */ public Solution(ListNode head) { this.head = head; this.current = head; r = new Random(); } /** Returns a random node's value. */ public int getRandom() { int res = head.val; ListNode n = head.next; int i = 2; while (n != null) { if (r.nextInt(i) == 0) { // For ith item, with probability 1/i, replace with new item. res = n.val; } n = n.next; i++; } return res; } } /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(head); * int param_1 = obj.getRandom(); */ }