Leetcode: Linked List Random Node

Posted 2020-08-21 neverlandly

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();

Solution 1: Reservior sampling: (wiki introduction)

Reservoir sampling is a family of randomized algorithms for randomly choosing a sample of k items from a list S containing n items, where n is either a very large or unknown number. Typically n is large enough that the list doesn‘t fit into main memory.

example: size = 1

Suppose we see a sequence of items, one at a time. We want to keep a single item in memory, and we want it to be selected at random from the sequence. If we know the total number of items (n), then the solution is easy: select an index i between 1 and n with equal probability, and keep the i-th element. The problem is that we do not always know n in advance. A possible solution is the following:

• Keep the first item in memory.
• When the i-th item arrives (for ):
  • with probability , keep the new item (discard the old one)
  • with probability , keep the old item (ignore the new one)

So:

• when there is only one item, it is kept with probability 1;
• when there are 2 items, each of them is kept with probability 1/2;
• when there are 3 items, the third item is kept with probability 1/3, and each of the previous 2 items is also kept with probability (1/2)(1-1/3) = (1/2)(2/3) = 1/3;
• by induction, it is easy to prove that when there are n items, each item is kept with probability 1/n.

 

This problem is size=1

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

 

解Size = k的问题见：https://discuss.leetcode.com/topic/53753/brief-explanation-for-reservoir-sampling

PROBLEM:

• Choose k entries from n numbers. Make sure each number is selected with the probability of k/n

BASIC IDEA:

• Choose 1, 2, 3, ..., k first and put them into the reservoir.
• For k+1, pick it with a probability of k/(k+1), and randomly replace a number in the reservoir.
• For k+i, pick it with a probability of k/(k+i), and randomly replace a number in the reservoir.
• Repeat until k+i reaches n

PROOF:

• For k+i, the probability that it is selected and will replace a number in the reservoir is k/(k+i)
• For a number in the reservoir before (let‘s say X), the probability that it keeps staying in the reservoir is
  • P(X was in the reservoir last time) × P(X is not replaced by k+i)
  • = P(X was in the reservoir last time) × (1 - P(k+i is selected and replaces X))
  • = k/(k+i-1) × （1 - k/(k+i) × 1/k）
  • = k/(k+i)
• When k+i reaches n, the probability of each number staying in the reservoir is k/n

EXAMPLE

• Choose 3 numbers from [111, 222, 333, 444]. Make sure each number is selected with a probability of 3/4
• First, choose [111, 222, 333] as the initial reservior
• Then choose 444 with a probability of 3/4
• For 111, it stays with a probability of
  • P(444 is not selected) + P(444 is selected but it replaces 222 or 333)
  • = 1/4 + 3/4*2/3
  • = 3/4
• The same case with 222 and 333
• Now all the numbers have the probability of 3/4 to be picked