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.
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?
Example:
// 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);
// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
solution.getRandom();
public class Solution {
/**
* Start... When we read the first node head, if the stream ListNode stops here,
* we can just return the head.val. The possibility is 1/1.
*
* When we read the second node, we can decide if we replace the result r or
* not. The possibility is 1/2. So we just generate a random number between 0
* and 1, and check if it is equal to 1. If it is 1, replace r as the value of
* the current node, otherwise we don't touch r, so its value is still the value
* of head.
*
* When we read the third node, now the result r is one of value in the head or
* second node. We just decide if we replace the value of r as the value of
* current node(third node). The possibility of replacing it is 1/3, namely the
* possibility of we don't touch r is 2/3. So we just generate a random number
* between 0 ~ 2, and if the result is 2 we replace r.
*
* We can continue to do like this until the end of stream ListNode.
*/
ListNode head = null;
Random randomGenerator = null;
public Solution(ListNode head) {
this.head = head;
this.randomGenerator = new Random();
}
/** Returns a random node's value. */
public int getRandom() {
ListNode result = null;
ListNode current = head;
for (int n = 1; current != null; n++) {
if (randomGenerator.nextInt(n) == 0) {
result = current;
}
current = current.next;
}
return result.val;
}
}