Random Linked List Node
Problem Statement
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.
Implement the Solution class:
Solution(ListNode head): Initializes the object with the integer array nums.
int getRandom(): Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be choosen.
Constraints:
- The number of nodes in the linked list will be in the range
[1, 10^4]. -10^4 <= Node.val <= 10^4- At most
10^4calls will be made togetRandom.
Linear space complexity solution
A naive solution is to store all nodes in an ArrayList and then randomly pick an item:
python3 class Solution: def init(self, head: ListNode): self.nodes = [] current = head while current: self.nodes.append(current.val) current = current.next
def getRandom(self) -> int:
return random.choice(self.nodes)
The getRandom method has O(1) running time. What if the LinkedList is very long? Then self.nodes will consume a lot of memory.
Constant space complexity solution I
We associate each node with a random number and pick the node associated with the maximum number:
python3 class Solution: def init(self, head: ListNode): self.head = head
def getRandom(self) -> int:
current_max = -1
val = None
current = self.head
while current:
ran = random.random()
if ran > current_max:
current_max = ran
val = current.val
current = current.next
return val
Constant space complexity solution II
Another constant space complexity solution uses Reservoir sampling. It’s a special case with k = 1.
python3 class Solution: def init(self, head: ListNode): self.head = head
def getRandom(self) -> int:
node = self.head
res = None
index = 0
while node:
current = random.randint(0, index)
if current == index:
res = node.val
index += 1
node = node.next
return res
One advantange of this solution is it can be easily extended to cases with k > 1. For example, one can be asked to generate 5 random nodes. See a clean implementation here.