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();
*/
}