Citlalli-Cedar

README.md

@@ -1,4 +1,3 @@
-# Linked-List
 
 Using object oriented design constructs, define a Node class and LinkedList class for a singly linked list. Each node has integer data value and a link to the next node. The linked list class has a head node and the following methods defined.
 

linked_list/linked_list.py

@@ -1,5 +1,9 @@
 
 # Defines a node in the singly linked list
+from ast import Delete
+import re
+
+
 class Node:
 
     def __init__(self, value, next_node = None):
@@ -9,94 +13,176 @@ def __init__(self, value, next_node = None):
 # 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:
+            return self.head.value
+        return None
 
     # 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
+
+        self.head = Node(value, 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(n)
+    # Space Complexity: O(1)
     def search(self, value):
-        pass
+        current_node = self.head
+        if current_node == None:
+            return False
+        while current_node:
+            if current_node.value == value:
+                return True
+            current_node = current_node.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
+        count = 0
+        current_node = self.head
+        while current_node != None:
+            current_node = current_node.next
+            count +=1
+        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: 0(1)
     def get_at_index(self, index):
-        pass
+        count =0
+        current_node = self.head
+
+        while current_node != None:
+            if count == index:
+                return current_node.value
+            count += 1
+            current_node = current_node.next
+        return None
+
 
     # 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
+        current_node = self.head
+        if current_node == None:
+            return None
+
+        while current_node.next:
+            current_node = current_node.next
+
+        return current_node.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(1)
     def add_last(self, value):
-        pass
+        if self.head == None:
+            self.add_first(value)
+        else:
+            new_node = Node(value)
+            current_node = self.head
+
+            while current_node.next:
+                current_node = current_node.next
+            current_node.next = new_node
+
+
 
     # 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(self):
-        pass
+        if self.head == None:
+            return None
+
+        current_node = self.head
+        max_value = self.head.value
+
+        while current_node:
+            if current_node.value > max_value:
+                max_value = current_node.value
+            current_node = current_node.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 not self.head:
+            return
+        elif self.head.value == value:
+            self.head = self.head.next
+            return True
+        else:
+            current = self.head
+            while current.next and current.next.value != value:
+                current = current.next
+            current.next = current.next.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
+        current_node = self.head
 
-        while current:
-            helper_list.append(str(current.value))
-            current = current.next
+        while current_node:
+            helper_list.append(str(current_node.value))
+            current_node = current_node.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 not self.head:
+            return
+
+        current_node = self.head
+        previous_node = None
+
+        while current_node:
+            next = current_node.next
+            current_node.next = previous_node
+            previous_node = current_node
+            current_node = next
+
+        self.head = previous_node
+
     ## Advanced/ Exercises
     # returns the value at the middle element in the singly linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)
+    # Space Complexity: O(1)
     def find_middle_value(self):
-        pass
+        if not self.head:
+            return None
+
+        return self.get_at_index(int(self.length() / 2))
+
 
     # find the nth node from the end and return its value
     # assume indexing starts at 0 while counting to n