题目描述：

LeetCode 382. Linked List Random Node

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();

题目大意：

给定一个单链表，返回链表中一个随机节点的值。每一个节点应该被等可能的抽取。

进一步思考：

如果链表很大并且长度未知呢？你可以不使用额外的空间高效地解决此问题吗？

解题思路：

蓄水池抽样（Reservoir Sampling）

蓄水池抽样算法的等概率性可以用数学归纳法证明：

I   当链表长度为1时，random.randint(0, 0)恒等于0，因此抽到第1个元素的概率为1

II  假设抽取前n个元素的概率相等，均为1/n

III 当抽取第n+1个元素时：

若random.randint(0, n)等于0，则返回值替换为第n+1个元素，其概率为1/(n+1)；

否则，抽取的依然是前n个元素，其概率为1/n * n/(n+1) = 1/(n+1)

Python代码：

import random
class Solution(object):

    def __init__(self, head):
        """
        @param head The linked list's head. Note that the head is guanranteed to be not null, so it contains at least one node.
        :type head: ListNode
        """
        self.head = head

    def getRandom(self):
        """
        Returns a random node's value.
        :rtype: int
        """
        cnt = 0
        head = self.head
        while head:
            if random.randint(0, cnt) == 0:
                ans = head.val
            head = head.next
            cnt += 1
        return ans


# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()

 

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