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linked_list.rb
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284 lines (263 loc) · 6.92 KB
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# Defines a node in the singly linked list
class Node
attr_reader :data # allow external entities to read value but not write
attr_accessor :next # allow external entities to read or write next node
def initialize(value, next_node = nil)
@data = value
@next = next_node
end
end
# Defines the singly linked list
class LinkedList
def initialize
@head = nil # keep the head private. Not accessible outside this class
end
# method to add a new node with the specific data value in the linked list
# insert the new node at the beginning of the linked list
# Time Complexity: O(1)
# Space Complexity: O(1)
def add_first(value)
new_node = Node.new(value, @head)
@head = new_node
end
# method to find if the linked list contains a node with specified value
# returns true if found, false otherwise
# Time Complexity: O(n)
# Space Complexity: O(n)
def search(value)
if @head == nil
return false
end
current = @head
while current
if current.data == value
return true
current == current.next
end
end
return false
end
# method to return the max value in the linked list
# returns the data value and not the node
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_max
if @head == nil
return nil
end
current = @head
max = current.data
while current
if current.data > max
max = current.data
end
current = current.next
end
return max
end
# method to return the min value in the linked list
# returns the data value and not the node
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_min
if @head == nil
return nil
end
current = @head
min = current.data
while current
if current.data < min
min = current.data
end
current = current.next
end
return min
end
# method that returns the length of the singly linked list
# Time Complexity: 0(n)
# Space Complexity: O(1)
def length
length = 0
current = @head
while current != nil
length += 1
current = current.next
end
return length
end
# method that returns the value at a given index in the linked list
# index count starts at 0
# returns nil if there are fewer nodes in the linked list than the index value
# Time Complexity: O(n)
# Space Complexity: O(1)
def get_at_index(index)
value = nil
count = 0
current = @head
while current
if count == index
value = current.data
break
end
current = current.next
count += 1
end
return value
end
# method to print all the values in the linked list
# Time Complexity: O(n)
# Space Complexity: O(n)
def visit
return nil if !@head
current = @head
while current
puts current.data
end
end
# method to delete the first node found with specified value
# Time Complexity: O(n)
# Space Complexity: O(1)
def delete(value)
return if !@head
if @head.data == value
@head = @head.next
else
current = @head
while current.next
if current.next.data == value
current.next = current.next.next
break
end
current = current.next
end
end
end
# method to reverse the singly linked list
# note: the nodes should be moved and not just the values in the nodes
# Time Complexity: O(n)
# Space Complexity: O(1)
def reverse
current = @head
previous = nil
next_node = nil
while current
next_node = current.next
current.next = previous
previous = current
current = next_node
end
@head = previous
end
## Advanced Exercises
# returns the value at the middle element in the singly linked list
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_middle_value
return nil if self.length == 0
middle = self.length / 2
middle_value = self.get_at_index(middle)
return middle_value
end
# find the nth node from the end and return its value
# assume indexing starts at 0 while counting to n
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_nth_from_end(n)
index = (self.length - 1) - n
value = self.get_at_index(index)
return value
end
# checks if the linked list has a cycle. A cycle exists if any node in the
# linked list links to a node already visited.
# returns true if a cycle is found, false otherwise.
# Time Complexity: O(n)
# Space Complexity: O(1)
def has_cycle
return false if !@head
first = @head
second = @head
while first.next.next
second = second.next
if first == second
return true
end
end
return false
end
# Additional Exercises
# returns the value in the first node
# returns nil if the list is empty
# Time Complexity: O(1)
# Space Complexity: O(1)
def get_first
if @head == nil
return nil
else
return @head.data
end
end
# method that inserts a given value as a new last node in the linked list
# Time Complexity: O(n)
# Space Complexity: O(1)
def add_last(value)
if self.length == 0
new_node = Node.new(value)
@head = new_node
else
current = @head
while current.next
current = current.next
end
new_node = Node.new(value)
current.next = new_node
end
end
# method that returns the value of the last node in the linked list
# returns nil if the linked list is empty
# Time Complexity: O(n)
# Space Complexity: O(1)
def get_last
value = nil
current = @head
while current
if current.next == nil
value = current.data
break
end
current = current.next
end
return value
end
# method to insert a new node with specific data value, assuming the linked
# list is sorted in ascending order
# Time Complexity: O(n)
# Space Complexity: O(1)
def insert_ascending(value)
if !@head
new_node = Node.new(value, @head)
@head = new_node
else
current = @head
while current
if current.data <= value && current.next.data >= value
new_node = Node.new(value, current.next)
current.next = new_node
break
end
current = current.next
end
end
end
# Helper method for tests
# Creates a cycle in the linked list for testing purposes
# Assumes the linked list has at least one node
def create_cycle
return if @head == nil # don't do anything if the linked list is empty
# navigate to last node
current = @head
while current.next != nil
current = current.next
end
current.next = @head # make the last node link to first node
end
end