Linked List Random Node_probability of linked list nodes store key values-优快云博客

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/106785029

本文介绍了一种使用Reservoir Sampling算法从未知长度的链表中高效选取随机节点的方法，确保每个节点被选中的概率相等，且不使用额外空间。

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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();

思路：reseverior sampling；取到random == 0，就update candidate；

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

    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
    private Random random;
    private ListNode head;
    public Solution(ListNode head) {
        this.head = head;
        this.random = new Random();
    }
    
    /** Returns a random node's value. */
    public int getRandom() {
        int count = 0;
        ListNode candidate = head; // candidate首先是第一个node;
        ListNode cur = head;
        while(cur != null) {
            if(random.nextInt(++count) == 0) {
                candidate = cur;
            }
            cur = cur.next;
        }
        return candidate.val;
    }
}

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