原题链接在这里：https://leetcode.com/problems/linked-list-random-node/

题目：

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();

题解：

First assign head value to res. count = 1.

Then while current node next != null, move cur to next,  pick random number within [0, ++count).

If it is equal to 0, update res.

Time Complexity: getRandom, O(n). n is length of list.

Space: O(1).

AC Java:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     ListNode head;
11     Random rand;
12     
13     /** @param head The linked list's head.
14         Note that the head is guaranteed to be not null, so it contains at least one node. */
15     public Solution(ListNode head) {
16         this.head = head;
17         rand = new Random();
18     }
19     
20     /** Returns a random node's value. */
21     public int getRandom() {
22         ListNode cur = head;
23         int res = cur.val;
24         int count = 1;
25         while(cur.next != null){
26             cur = cur.next;
27             if(rand.nextInt(++count) == 0){
28                 res = cur.val;
29             }
30         }
31         
32         return res;
33     }
34 }
35 
36 /**
37  * Your Solution object will be instantiated and called as such:
38  * Solution obj = new Solution(head);
39  * int param_1 = obj.getRandom();
40  */

类似Random Pick Index.