LeetCode #382. Linked List Random Node

LawFile

于 2019-11-29 10:28:33 发布

阅读量151

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/103305825

LeetCode 专栏收录该内容

625 篇文章

订阅专栏

本文介绍了一种从链表中高效获取随机节点值的方法，确保每个节点被选中的概率相同，适用于链表长度未知的大规模数据集，通过遍历链表并使用随机数生成器实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

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();

class Solution {
public:
    /** @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) {
        l=head;
    }
    
    /** Returns a random node's value. */
    int getRandom() {
        int result=l->val;
        int n=1;
        ListNode* p=l;
        while(p)
        {
            int x=rand()%n;
            if(x==0) result=p->val;
            n++;
            p=p->next;
        }
        return result;
    }
    
private:
    ListNode* l;
};