0382. Linked List Random Node (M)

Linked List Random Node (M)

題目

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();

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題意

思路

最簡單的方法是直接遍歷鏈表，把值所有存到List中，最後隨機下標便可。

官方解答還有一種神奇的水塘採樣方法。

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代碼實現

Java

List存儲

class Solution {
    private List<Integer> list = new ArrayList<>();
    private Random random = new Random();

    /** @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) {
        while (head != null) {
            list.add(head.val);
            head = head.next;
        }
    }

    /** Returns a random node's value. */
    public int getRandom() {
        int index = random.nextInt(list.size());
        return list.get(index);
    }
}

水塘採樣

class Solution {
    private 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() {
        ListNode chosen = null;
        int count = 1;
        ListNode cur = head;

        while (cur != null) {
            if (Math.random() < 1.0 / count) {
                chosen = cur;
            }
            count++;
            cur = cur.next;
        }

        return chosen.val;
    }
}