Spruce-C16 | Zandra | LinkedList

linked_list/linked_list.py

@@ -2,99 +2,215 @@
 # Defines a node in the singly linked list
 class Node:
 
-    def __init__(self, value, next_node = None):
+    def __init__(self, value, next_node=None):
         self.value = value
         self.next = next_node
 
 # Defines the singly linked list
+
+
 class LinkedList:
     def __init__(self):
-      self.head = None # keep the head private. Not accessible outside this class
+        self.head = None  # keep the head private. Not accessible outside this class
 
     # returns the value in the first node
     # returns None if the list is empty
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: (O)1?
+    # Space Complexity: O(1)?
+
     def get_first(self):
-        pass
+        if self.head == None:
+            return None
 
+        return self.head.value
 
     # 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: ?
+    # Time Complexity: (O)1?
+    # Space Complexity: O(1)?
+
     def add_first(self, value):
-        pass
+        # This method adds a new node with the given value to the head of the list
+        new_node = Node(value)
+        new_node.next = self.head
+        self.head = new_node
 
     # method to find if the linked list contains a node with specified value
     # returns true if found, false otherwise
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)?
+    # Space Complexity: O(1)?
+
     def search(self, value):
-        pass
+        # This method returns True or False if the list contains the given value.
+        if self.head == None:
+            return False
+
+        current = self.head
+
+        while current != None:
+            if current.value == value:
+                return True
+            current = current.next
+
+        return False
 
     # method that returns the length of the singly linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)?
+    # Space Complexity: O(1)?
+
     def length(self):
-        pass
+        # This method returns the size of the list.
+        current = self.head
+        count = 0
+
+        while current != None:
+            count += 1
+            current = current.next
+
+        return count
 
     # method that returns the value at a given index in the linked list
     # index count starts at 0
     # returns None if there are fewer nodes in the linked list than the index value
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)?
+    # Space Complexity: O(1)?
+
     def get_at_index(self, index):
-        pass
+        # This method returns the value of the node at the given index. It returns None if the list does not have that many elements.
+        if index < 0:
+            return None
+
+        current_index = 0
+        current = self.head
+
+        while current is not None and current_index < index:
+            current = current.next
+            current_index += 1
+
+        if current is None:
+            return None
+
+        return current.value
 
     # method that returns the value of the last node in the linked list
     # returns None if the linked list is empty
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n) ?
+    # Space Complexity: O(1)?
+
     def get_last(self):
-        pass
+        # This method returns the value of the last node in the list.
+        if self.head == None:
+            return None
+
+        current = self.head
+
+        while current.next != None:
+            current = current.next
+
+        return current.value
 
     # method that inserts a given value as a new last node in the linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)?
+    # Space Complexity: O(n)?
+
     def add_last(self, value):
-        pass
+        # This method adds a new node to the rear of the list.
+        if self.head == None:
+            new_node = Node(value)
+            self.head = new_node
+            return
+
+        current = self.head
+
+        while current.next != None:
+            current = current.next
+
+        # new_node = Node(value)
+        # current.next = new_node
+        current.next = Node(value)
 
     # method to return the max value in the linked list
     # returns the data value and not the node
+
     def find_max(self):
-        pass
+        # This method finds the largest value in the list, assuming you can use >, or < to compare each element in the list.
+        if self.head == None:
+            return None
+
+        current = self.head
+        max_value = -100
+
+        while current != None:
+            if current.value > max_value:
+                max_value = current.value
+
+            current = current.next
+
+        return max_value
 
     # method to delete the first node found with specified value
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)?
+    # Space Complexity: O(1)?
     def delete(self, value):
-        pass
+
+        # If the first node is the one we want to delete, that's a
+        # special case.
+        if self.head and self.head.value == value:
+            self.head = self.head.next
+            return
+
+        # Now iterate through the rest of the list and check
+        # for the target value at the node _after_ the current one.
+        current = self.head
+        while current:
+            if current.next and current.next.value == value:
+                current.next = current.next.next
+                return
+
+            current = current.next
 
     # method to print all the values in the linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)?
+    # Space Complexity: O(n)?
+
     def visit(self):
         helper_list = []
         current = self.head
 
         while current:
             helper_list.append(str(current.value))
             current = current.next
-        
+
         print(", ".join(helper_list))
 
     # 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: ?
+    # Time Complexity: O(n) ?
+    # Space Complexity: O(1)?
+
     def reverse(self):
-        pass
-
+        # If the list is empty or has only one element, then there is nothing to do
+        if not self.head or not self.head.next:
+            return
+
+        # The list is of length AT LEAST 2. Time to reverse
+        # the pointers
+        p1 = self.head
+        p2 = self.head.next
+        p1.next = None  # this is now the end of the new list
+        while p2:
+            temp = p2.next
+            p2.next = p1  # p2 now points at the previous node
+            p1 = p2  # advance p1 to the next node (in the old ordering)
+            p2 = temp
+
+        self.head = p1
+
     ## Advanced/ Exercises
     # returns the value at the middle element in the singly linked list
     # Time Complexity: ?
     # Space Complexity: ?
+
     def find_middle_value(self):
         pass
 
@@ -125,4 +241,4 @@ def create_cycle(self):
         while current.next != None:
             current = current.next
 
-        current.next = self.head # make the last node link to first node
+        current.next = self.head  # make the last node link to first node