382. Linked List Random Node

https://leetcode.com/problems/linked-list-random-node/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.

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();

Thoughts

满足Uniform distribution的随机采链表节点值，要求O(1)额外空间。O(1)空间采样=>蓄水池，遍历并根据当前长度做采样，当采到上次采过的元素则做替换。

Code

/*
 * @lc app=leetcode id=382 lang=cpp
 *
 * [382] Linked List Random Node
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *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. */
    Solution(ListNode* head) {
        head_ = head;
    }
    
    /** Returns a random node's value. */
    int getRandom() {
        int res = -1;
        auto cur = head_;
        for (int cnt = 1; cur != nullptr; cur = cur->next, ++cnt) {
            if (rand() % cnt == 0) res = cur->val;
        }
        return res;
    }
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution* obj = new Solution(head);
 * int param_1 = obj->getRandom();
 */
// @lc code=end

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    ListNode 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. */
    public Solution(ListNode head) {
        this.head = head;
    }

    /** Returns a random node's value. */
    public int getRandom() {
        int res = -1;
        Random rand = new Random();
        int count = 0;
        ListNode node = head;
        while (node != null) {
            if (rand.nextInt(++count) == 0) {
                res = node.val;
            }
            node = node.next;
        }
        return res;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */

Analysis

每次getRandom 需要O(N).