Leetcode: Linked List Random Node

Linked List Random Node

Similar Problems:

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?


// 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.

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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