Spruce - Angela Fan

linked_list/linked_list.py

@@ -1,75 +1,181 @@
 
 # Defines a node in the singly linked list
+# from ast import Pass
+
+
 class Node:
 
     def __init__(self, value, next_node = None):
         self.value = value
         self.next = next_node
+        # self.previous = previous_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
+        # self.length = 0 # keep the length private. Not accessible outside this class
+        # if self.head == None:
+        #     self.head = Node(None)
+        #     self.length = 0
+        # else:
+        #     self.length = 1
 
     # 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
+        else: 
+            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
+
+        if self.head == None:
+            self.head = Node(value)
+        else:
+            new_node = Node(value)
+            new_node.next = self.head
+            self.head = new_node
+            return self.head
+
+        # if self.head == None:
+        #     return None 
+
+        # self.head = Node(value, self.head)
+        # self.length += 1
+        # return self.head
 
     # 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(1)
+    # Space Complexity: O(1)
     def search(self, value):
-        pass
+        if self.head == None:
+            return False
+
+        if self.head.value == value:
+            return True
+
+        current = self.head
+        while current:
+            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
+
+        length_counter = 0
+        current = self.head
+        while (current != None):
+            length_counter += 1
+            current = current.next
+
+        return length_counter
+
 
     # 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
+
+        # if self.head == None:
+        #     return None 
+
+        if self.length() <= index:
+            return None
+
+        current_value = 0 
+        current_node = self.head
+        while current_node:
+            if current_value == index:
+                return current_node.value
+            current_node = current_node.next
+            current_value += 1
+
+        return current_node.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
+        if self.head == None:
+            return None
+
+        current = self.head
+        while current.next:
+            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(1)
+    # Space Complexity: O(1)
     def add_last(self, value):
-        pass
+        if self.head == None:
+            self.head = Node(value)
+        else:
+            current = self.head
+            while current.next:
+                current = current.next
+            current.next = Node(value)
+
+            return self.head
 
     # method to return the max value in the linked list
     # returns the data value and not the node
     def find_max(self):
-        pass
+        if self.head == None:
+            return None
+
+        current = self.head
+        max_value = current.value
+        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 self.head == None:
+            return None
+
+        current = self.head
+        previous = None
+        while current:
+            if current.value == value:
+                if previous == None:
+                    self.head = current.next
+                    return
+                else:
+                    previous.next = current.next
+                    return
+            else:
+                previous = current
+                current = current.next
+
+        return None
 
     # method to print all the values in the linked list
     # Time Complexity: ?
@@ -89,40 +195,81 @@ def visit(self):
     # Time Complexity: ?
     # Space Complexity: ?
     def reverse(self):
-        pass
+        if self.head == None:
+            return None
+
+        previous = None
+        current = self.head
+
+        while current != None:
+            next_node = current.next
+            current.next = previous
+            previous = current
+            current = next_node
+        self.head = previous
+        return previous
 
     ## Advanced/ Exercises
     # returns the value at the middle element in the singly linked list
     # Time Complexity: ?
     # Space Complexity: ?
     def find_middle_value(self):
-        pass
+        if self.head == None:
+            return None
+
+        current = self.head
+        middle_value = None
+        for i in range(self.length // 2):
+            current = current.next
+            middle_value = current.value
+
+        return middle_value
 
     # 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(self, n):
-        pass
+        if self.head == None:
+            return None
+
+        current = self.head
+        for i in range(n):
+            current = current.next
+
+        return current.value
 
     # 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(self):
-        pass
+        if self.head == None:
+            return None
+
+        current = self.head
+        visited = []
+        while current:
+            if current in visited: 
+                return True
+            else:
+                visited.append(current)
+                current = current.next
+        return False
 
     # 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(self):
         if self.head == None:
-            return
+            return None
 
         # navigate to last node
         current = self.head
         while current.next != None:
             current = current.next
 
         current.next = self.head # make the last node link to first node
+
+        return self.head