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Ports - Jessica #19

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352 changes: 238 additions & 114 deletions lib/linked_list.rb
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Original file line number Diff line number Diff line change
Expand Up @@ -12,152 +12,276 @@ def initialize(value, next_node = nil)

# Defines the singly linked list
class LinkedList
def initialize
@head = nil # keep the head private. Not accessible outside this class
end
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:
# Space Complexity
def add_first(value)
raise NotImplementedError
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)
@head = Node.new(value, @head) # i don't fully understand why next_node is @head
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You're setting next_node to the old value of @head. So the new node links to the start of the old list.


# method to find if the linked list contains a node with specified value
# returns true if found, false otherwise
# Time Complexity:
# Space Complexity
def search(value)
raise NotImplementedError
end
return @head.data
end

# method to return the max value in the linked list
# returns the data value and not the node
# Time Complexity:
# Space Complexity
def find_max
raise NotImplementedError
end
# method to find if the linked list contains a node with specified value
# returns true if found, false otherwise
# Time Complexity: O(n), where n is the length of the list
# Space Complexity: O(1)
def search(value)
current = @head

# method to return the min value in the linked list
# returns the data value and not the node
# Time Complexity:
# Space Complexity
def find_min
raise NotImplementedError
while current
return true if current.data == value
current = current.next
end

return false
end

# method that returns the length of the singly linked list
# Time Complexity:
# Space Complexity
def length
raise NotImplementedError
end
# method to return the max value in the linked list
# returns the data value and not the node
# Time Complexity: O(n), where n is the length of the list
# Space Complexity: O(1)
def find_max
current = @head

# 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:
# Space Complexity
def get_at_index(index)
raise NotImplementedError
end
return nil unless current

# method to print all the values in the linked list
# Time Complexity:
# Space Complexity
def visit
raise NotImplementedError
end
max_value = current.data

# method to delete the first node found with specified value
# Time Complexity:
# Space Complexity
def delete(value)
raise NotImplementedError
while current
max_value = current.data > max_value ? current.data : max_value
current = current.next
end

# method to reverse the singly linked list
# note: the nodes should be moved and not just the values in the nodes
# Time Complexity:
# Space Complexity
def reverse
raise NotImplementedError
end
return max_value
end

# method to return the min value in the linked list
# returns the data value and not the node
# Time Complexity: O(n), where n is the length of the list
# Space Complexity: O(1)
def find_min
current = @head

## Advanced Exercises
# returns the value at the middle element in the singly linked list
# Time Complexity:
# Space Complexity
def find_middle_value
raise NotImplementedError
return nil unless current

min_value = current.data

while current
min_value = current.data < min_value ? current.data : min_value
current = current.next
end

# find the nth node from the end and return its value
# assume indexing starts at 0 while counting to n
# Time Complexity:
# Space Complexity
def find_nth_from_end(n)
raise NotImplementedError
return min_value
end

# method that returns the length of the singly linked list
# Time Complexity: O(n), where n is the length of the list
# Space Complexity: O(1)
def length
return 0 unless @head

length = 0

current = @head
while current
current = current.next
length += 1
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:
# Space Complexity
def has_cycle
raise NotImplementedError
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), where n is the value of the index
# Space Complexity: O(1)
def get_at_index(index)
return nil unless @head

current = @head

index.times do
current = current.next

return nil unless current
end

return current.data
end

# method to print all the values in the linked list
# Time Complexity: O(n), where n is the length of the list
# Space Complexity: O(1), unless printing takes space?
def visit
return nil unless @head

# Additional Exercises
# returns the value in the first node
# returns nil if the list is empty
# Time Complexity:
# Space Complexity
def get_first
raise NotImplementedError
current = @head

while current
puts current.data
current = current.next
end
end

# method that inserts a given value as a new last node in the linked list
# Time Complexity:
# Space Complexity
def add_last(value)
raise NotImplementedError
# method to delete the first node found with specified value
# Time Complexity: O(n), where n is the length of the list
# Space Complexity: O(1)
def delete(value)
# if the value doesn't exist, just return the list
return self unless self.search(value)

# delete the value if at the head
if @head.data == value
@head = @head.next ? @head.next : nil
return
end

# method that returns the value of the last node in the linked list
# returns nil if the linked list is empty
# Time Complexity:
# Space Complexity
def get_last
raise NotImplementedError
# delete value somewhere else
previous = @head
current = @head.next
while current
if current.data == value
previous.next = current.next ? current.next : nil
end
current = current.next
previous = previous.next
end

# method to insert a new node with specific data value, assuming the linked
# list is sorted in ascending order
# Time Complexity:
# Space Complexity
def insert_ascending(value)
raise NotImplementedError
return self
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), where n is the length of the list
# Space Complexity: O(1)?
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Since you're reversing in place yes it's O(1) in space complexity.

def reverse
return nil unless @head
return self if self.length < 2

temporary = nil
previous = nil
current = @head
puts self.visit

# printing things to try to trace what's happening
while current
puts "------------"
puts "temporary: #{temporary.data if temporary}"
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You can remove these when you submit.

puts "previous: #{previous.data if previous}"
puts "current: #{current.data}"
puts "next: #{current.next.data if current.next}"
puts "list: #{self.visit}"
temporary = current.next
current.next = previous
previous = current
current = temporary

# puts "temporary: #{temporary.data if temporary}"
# puts "previous: #{previous.data}"
# puts "current: #{current.data if current}"
# puts "next: #{current.next.data if current && current.next}"
end

@head = previous

puts self.visit
return self
end

## Advanced Exercises
# returns the value at the middle element in the singly linked list
# Time Complexity:
# Space Complexity
def find_middle_value
raise NotImplementedError
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-m), where n is the length of the list and m is the index
# Space Complexity: O(1)
def find_nth_from_end(n)
return nil if self.length - 1 < n

self.get_at_index(self.length - 1 - n)
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Very good use of existing methods!

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
# 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:
# Space Complexity
def has_cycle
raise NotImplementedError
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
return @head ? @head.data : nil
end

# navigate to last node
# method that inserts a given value as a new last node in the linked list
# Time Complexity: O(n), where n is the length of the list
# Space Complexity: O(1)
def add_last(value)
if @head
current = @head
while current.next != nil
current = current.next

while current.next
current = current.next
end

current.next = @head # make the last node link to first node
return current.next = Node.new(value)
else
self.add_first(value)

return @head
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), where n is the length of the list
# Space Complexity: O(1)
def get_last
return nil unless @head

current = @head
current = current.next while current.next

return current.data
end

# method to insert a new node with specific data value, assuming the linked
# list is sorted in ascending order
# Time Complexity:
# Space Complexity
def insert_ascending(value)
raise NotImplementedError
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
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