Linked List Random Node
- 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.
What if the linked list is extremely large and its length is unknown to you? Could you solve this efficiently without using extra space?
// 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();
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()
Original URL: https://code.dennyzhang.com/linked-list-random-node