leetcode:linked List Random Node

Given a singly linked list,returnA random node ' s value from the linked list. Each node must has the same probability of being chosen.Follow Up:whatifThe linked list is extremely large and its length are unknown to you? Could You Solve ThisEfficiently without using extra space?Example://Init a singly linked list [I/b].ListNode head =NewListNode (1); Head.next=NewListNode (2); Head.next.next=NewListNode (3); Solution Solution=Newsolution (head);//Getrandom () should return either 1, 2, or 3 randomly. Each element should has equal probability of returning.Solution.getrandom ();

Solution 1:reservior Sampling: (wiki introduction)

Reservoir sampling is a family of randomized algorithms in 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, the list doesn ' t fit into main memory.

Example:size = 1

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

• Keep the first item in memory.
• When the i-th item is 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 was only one item, it was kept with probability 1;
• When there was 2 items, each of them was kept with probability 1/2;
• When there was 3 items, the third item is kept with probability 1/3, and all of the previous 2 items is also kept with PR Obability (1/2) (1-1/3) = (1/2) (2/3) = 1/3;
• By induction, it's easy-to-prove that's when there was n items, each item was 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 classSolution {Ten ListNode start; One  A     /** @paramhead the linked list ' s head. - Note that the head was guaranteed to was not null and so it contains at least one node.*/ -      Publicsolution (ListNode head) { the          This. Start =head; -     } -      -     /**Returns A random node ' s value.*/ +      Public intGetrandom () { -Random random =NewRandom (); +ListNode cur =start; A         intval =Start.val; at          -          for(intI=1; cur!=NULL; i++) { -             if(Random.nextint (i) = = 0) { -val =Cur.val; -             } -Cur =Cur.next; in         } -         returnVal; to     } + } -  the /** * * Your Solution Object would be instantiated and called as such: $ * Solution obj = new solution (head);Panax Notoginseng * int param_1 = Obj.getrandom (); -  */

To solve the problem of size = k See: https://discuss.leetcode.com/topic/53753/brief-explanation-for-reservoir-sampling

Problem:

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

BASIC idea:

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

PROOF:

• k+ifor, the probability. It is selected and would replace a number in the reservoir isk/(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)XP(X is not replaced by k+i)
  • = P(X was in the reservoir last time) x ( 1 - P(k+i is selected and replaces X) )
  • = k/(k+i-1) x ( 1 - k/(k+i) x 1/k )
  • =k/(k+i)
• When k+i reaches n , the probability of each number staying in the reservoir arek/n

EXAMPLE

• Choose 3 numbers from [111, 222, 333, 444] . Make sure each number are selected with a probability of3/4
• First, choose as the [111, 222, 333] initial reservior
• Then choose with 444 a probability of3/4
• 111for, 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 and 222333
• Now all the numbers has the probability of to be 3/4 picked

leetcode:linked List Random Node