382. Linked List Random Node

1. Description

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.

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?

2. Example

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.

3. Constraints

• The number of nodes in the linked list will be in the range [1, 10$^4$].
• -10$^4$ <= Node.val <= 10$^4$
• At most 10$^4$ calls will be made to getRandom.

4. Solutions

Reservoir Sampling

n is the number of nodes in head
Solution(head): O(1) time, O(1) space
getRandom(): O(n) time, O(1) space

class Solution {
public:
    Solution(ListNode *head) {
        this->head = head;
    }

    int getRandom() {
        ListNode *node = head;
        int result;
        int count = 0;

        while (node != nullptr) {
            ++count;

            if (rand() % count == 0) {
                result = node->val;
            }

            node = node->next;
        }

        return result;
    }

private:
    ListNode *head = nullptr;
};