Add Find the middle element of a linked list Algorithm

Author: th92rodr

google_interview_prep/find_middle_element_of_linked_list.py

@@ -0,0 +1,88 @@
+'''
+Find the middle element of a linked list
+
+Our challenge is to now find the middle node in a linked list.
+We don't initially know the length of a linked list, all we have is a single node which acts as the head of the linked list and which we can access all other nodes by traversing through each nodes "next" property.
+We can continuously loop through each node until we get to a node that has a "next" property of null, which means we have reached the last node.
+
+https://coderbyte.com/algorithm/linked-list-middle-element
+'''
+
+
+class Node:
+    def __init__(self, data):
+        self.data = data
+        self.next = None
+
+
+class LinkedList:
+    def __init__(self):
+        self.root = None
+
+    def insert_at_end(self, data):
+        new_node = Node(data)
+
+        if self.root is None:
+            self.root = new_node
+        else:
+            current = self.root
+            while current.next is not None:
+                current = current.next
+            current.next = new_node
+
+    def print_list(self):
+        print('List:')
+        current = self.root
+        while current is not None:
+            print(current.data, '-> ', end='')
+            current = current.next
+
+    '''
+    Traverse all the list nodes and count how many elements there are in total.
+    Then traverse it again this time stopping at the total/2 node.
+    '''
+    def get_middle_element_1(self):
+        current = self.root
+        length = 0
+
+        # walk until the end of the list to find it's length
+        while current is not None:
+            length += 1
+            current = current.next
+
+        current = self.root
+        # walk until the middle of the list length
+        for i in range(1, length//2):
+            current = current.next
+
+        return current.data
+
+    '''
+    Traverse the list using two pointers:
+    one that will traverse the linked list one node at a time,
+    and the other pointer will traverse two nodes at a time.
+    This way when the faster pointer reaches the end of the linked list,
+    the slower pointer will be halfway there.
+    '''
+    def get_middle_element_2(self):
+        fastPointer = self.root.next
+        slowPointer = self.root
+
+        # the fast pointer walks two nodes at a time and the slow just one
+        while fastPointer.next is not None and fastPointer.next.next is not None:
+            fastPointer = fastPointer.next.next
+            slowPointer = slowPointer.next
+
+        return slowPointer.data
+
+
+list = LinkedList()
+
+for x in range(1, 11):
+    list.insert_at_end(x)
+
+list.print_list()
+
+print('\n\nMiddle list element:', list.get_middle_element_1())
+
+print('\n\nMiddle list element:', list.get_middle_element_2())