README.md

382. Linked List Random Node

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();

Solutions (Python)

1. Reservoir Sampling

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:

    def __init__(self, head: ListNode):
        """
        @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node.
        """
        self.head = head


    def getRandom(self) -> int:
        """
        Returns a random node's value.
        """
        curr = self.head
        cnt = 0
        ret = 0

        while curr:
            cnt += 1
            if random.randint(1, cnt) == cnt:
                ret = curr.val
            curr = curr.next

        return ret


# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()