用于从单链表中选择随机节点的Python程序
给定一个单链表,从链表中随机选择一个节点(如果列表中有 N 个节点,选择一个节点的概率应该是 1/N)。您将获得一个随机数生成器。
下面是一个简单的解决方案:
- 通过遍历列表来计算节点数。
- 再次遍历列表并选择概率为 1/N 的每个节点。可以通过为第 i 个节点生成一个从 0 到 Ni 的随机数来完成选择,并且仅当生成的数字等于 0(或从 0 到 Ni 的任何其他固定数字)时才选择第 i 个节点。
通过上述方案,我们得到了统一的概率。
i = 1, probability of selecting first node = 1/N
i = 2, probability of selecting second node =
[probability that first node is not selected] *
[probability that second node is selected]
= ((N-1)/N)* 1/(N-1)
= 1/N
类似地,其他选择其他节点的概率为 1/N
上述方案需要对链表进行两次遍历。
如何选择一个只允许遍历一次的随机节点?
这个想法是使用 Reservoir Sampling。以下是步骤。这是 Reservoir Sampling 的一个更简单的版本,因为我们只需要选择一个键而不是 k 个键。
(1) Initialize result as first node
result = head->key
(2) Initialize n = 2
(3) Now one by one consider all nodes from 2nd node onward.
(a) Generate a random number from 0 to n-1.
Let the generated random number is j.
(b) If j is equal to 0 (we could choose other fixed numbers
between 0 to n-1), then replace result with the current node.
(c) n = n+1
(d) current = current->next
下面是上述算法的实现。
Python
# Python program to randomly select a
# node from singly linked list
import random
# Node class
class Node:
# Constructor to initialize the
# node object
def __init__(self, data):
self.data= data
self.next = None
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
# A reservoir sampling-based function
# to print a random node from a
# linked list
def printRandom(self):
# If list is empty
if self.head is None:
return
if self.head and not self.head.next:
print "Randomly selected key is %d" %(self.head.data)
# Use a different seed value so that we don't get
# same result each time we run this program
random.seed()
# Initialize result as first node
result = self.head.data
# Iterate from the (k+1)th element nth element
# because we iterate from (k+1)th element, or
# the first node will be picked more easily
current = self.head.next
n = 2
while(current is not None):
# Change result with probability 1/n
if (random.randrange(n) == 0 ):
result = current.data
# Move to next node
current = current.next
n += 1
print "Randomly selected key is %d" %(result)
# Function to insert a new node at
# the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Utility function to print the linked
# LinkedList
def printList(self):
temp = self.head
while(temp):
print temp.data,
temp = temp.next
# Driver code
llist = LinkedList()
llist.push(5)
llist.push(20)
llist.push(4)
llist.push(3)
llist.push(30)
llist.printRandom()
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
请注意,上述程序是基于随机函数的结果,可能会产生不同的输出。
这是如何运作的?
让列表中总共有 N 个节点。从最后一个节点开始更容易理解。
最后一个节点是结果的概率只是 1/N [对于最后一个或第 N 个节点,我们生成一个介于 0 到 N-1 之间的随机数,如果生成的数字为 0(或任何其他固定值),则将最后一个节点作为结果数字]
倒数第二个节点是结果的概率也应该是 1/N。
The probability that the second last node is result
= [Probability that the second last node replaces result] X
[Probability that the last node doesn't replace the result]
= [1 / (N-1)] * [(N-1)/N]
= 1/N
同样,我们可以显示第三个最后一个节点和其他节点的概率。
有关详细信息,请参阅有关从单链表中选择随机节点的完整文章!