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Linked_List_Random_Node.py
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Linked_List_Random_Node.py
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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();
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
# My Solution
# O(n) Time and O(n) Space
import random
class Solution:
def __init__(self, head: ListNode):
"""
@param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node.
"""
self.llist = []
start = head
while start:
self.llist.append(start.val)
start = start.next
def getRandom(self) -> int:
"""
Returns a random node's value.
"""
return random.choice(self.llist)
# To address the follow-up question, we can use Reservoir Sampling
# O(n) Time and O(1) Space
class Solution:
def __init__(self, head: ListNode):
"""
@param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node.
"""
self.head = head
def getRandom(self) -> int:
"""
Returns a random node's value.
"""
scope = 1
chosen_value = 0
curr = self.head
while curr:
# decide whether to include the element in reservoir
if random.random() < 1 / scope:
chosen_value = curr.val
# move on to the next node
curr = curr.next
scope += 1
return chosen_value
# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()