LeetCode #382: Linked List Random Node

Problem Statement

(Problem Link) 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();

Analysis

Reservoir sampling

Reservoir Sampling Sample 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 i>1):
  
  • with probability 1/i, keep the new item instead of the current item; or equivalently
  • with probability 1-1/i, keep the current item and discard the new item.

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.

Solution

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def __init__(self, head):        """        @param head The linked list's head.        Note that the head is guaranteed to be not null, so it contains at least one node.        :type head: ListNode        """        self.head = head    def getRandom(self):        """        Returns a random node's value.        :rtype: int        """        from random import randint        p = self.head        res = p.val        i = 2        p = p.next        while p:            if randint(1, i) == 1:                res = p.val            p = p.next            i += 1        return res# Your Solution object will be instantiated and called as such:# obj = Solution(head)# param_1 = obj.getRandom()