Linked List Random Node

Similar Problems:

- Tag: #reservoirsampling

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();

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/linked-list-random-node ## Basic Ideas: Reservoir Sampling ## Choose k element from a huge list without knowing the length ## ## Complexity: Time O(n), Space O(1) ## # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None import random class 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 """ if self.head is None: return None if self.head.next is None: return self.head.val res = self.head.val p = self.head.next i = 2 while p: # Keep with possibility of 1/i random_v = random.randint(1, i) # Why we won't quit? if random_v == 1: res = p.val p = p.next i = i+1 return res # Your Solution object will be instantiated and called as such: # obj = Solution(head) # param_1 = obj.getRandom()

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