LeetCode 382. Linked List Random Node

題目描述：

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();

題目大意：

給定一個單鏈表，返回鏈表中一個隨機節點的值。每一個節點應該被等可能的抽取。

進一步思考：

如果鏈表很大並且長度未知呢？你可以不使用額外的空間高效地解決此問題嗎？

解題思路：

蓄水池抽樣（Reservoir Sampling）

蓄水池抽樣算法的等概率性可以用數學歸納法證明：

I   當鏈表長度爲1時，random.randint(0, 0)恆等於0，因此抽到第1個元素的概率爲1

II  假設抽取前n個元素的概率相等，均爲1/n

III 當抽取第n+1個元素時：

若random.randint(0, n)等於0，則返回值替換爲第n+1個元素，其概率爲1/(n+1)；

否則，抽取的依然是前n個元素，其概率爲1/n * n/(n+1) = 1/(n+1)

Python代碼：

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
      
 本文鏈接：http://bookshadow.com/weblog/2016/08/10/leetcode-linked-list-random-node/
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