0382. Linked List Random Node (M)
阿新 • • 發佈:2020-12-02
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();
題意
思路
最簡單的方法是直接遍歷連結串列,把值全部存到List中,最後隨機下標即可。
官方解答還有一種神奇的水塘取樣方法。
程式碼實現
Java
List儲存
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); } }
水塘取樣
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;
}
}