Create 876_1_Middle_of_Linked_List.py

Brute force approach

Author: preetisharma0123

876_1_Middle_of_Linked_List.py

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+""""
+876. Middle of the Linked List
+
+Given the head of a singly linked list, return the middle node of the linked list.
+
+If there are two middle nodes, return the second middle node."""
+
+#Brute-force approach-1
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        curr = head
+        count = 1
+
+        while curr.next:
+            curr = curr.next
+            count+=1
+
+        curr = head
+        if count%2 == 0:
+            for count in range(count,(count//2),-1):
+                curr = curr.next
+                
+            
+        else:
+            for count in range(count,(count//2)+1,-1):
+                curr = curr.next
+                
+        
+        head = curr
+        return head
+
+
+
+
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+
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+