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.

Implement the Solution class:

• Solution(ListNode head) Initializes the object with the head of the singly-linked list head.
• int getRandom() Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be chosen.

Example 1:

Input
["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 3, 2, 2, 3]

Explanation
Solution solution = new Solution([1, 2, 3]);
solution.getRandom(); // return 1
solution.getRandom(); // return 3
solution.getRandom(); // return 2
solution.getRandom(); // return 2
solution.getRandom(); // return 3
// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.

Constraints:

• The number of nodes in the linked list will be in the range [1, 104].
• -104 <= Node.val <= 104
• At most 104 calls will be made to getRandom.

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?

SOLUTION:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

import random

class Solution:

    def __init__(self, head: Optional[ListNode]):
        self.head = head

    def getRandom(self) -> int:
        res = self.head.val
        n = 2
        curr = self.head.next
        while curr:
            isSwap = random.randrange(0, n)
            if isSwap == 0:
                res = curr.val
            curr = curr.next
            n += 1
        return res


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