Delete the Middle Node of a Linked List

Author: shashankvishwakarma

src/main/java/leetcode/medium/DeleteMiddleNodeOfLinkedList.java

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+package leetcode.medium;
+
+import dsa.linkedlist.ListNode;
+
+/**
+ * https://leetcode.com/problems/delete-the-middle-node-of-a-linked-list/
+ * <p>
+ * Example 1:
+ * Input: head = [1,3,4,7,1,2,6]
+ * Output: [1,3,4,1,2,6]
+ * Explanation:
+ * The above figure represents the given linked list. The indices of the nodes are written below.
+ * Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
+ * We return the new list after removing this node.
+ * <p>
+ * Example 2:
+ * Input: head = [1,2,3,4]
+ * Output: [1,2,4]
+ * Explanation:
+ * The above figure represents the given linked list.
+ * For n = 4, node 2 with value 3 is the middle node, which is marked in red.
+ * <p>
+ * Example 3:
+ * Input: head = [2,1]
+ * Output: [2]
+ * Explanation:
+ * The above figure represents the given linked list.
+ * For n = 2, node 1 with value 1 is the middle node, which is marked in red.
+ * Node 0 with value 2 is the only node remaining after removing node 1.
+ */
+class DeleteMiddleNodeOfLinkedList {
+
+    public static void main(String[] args) {
+        ListNode head = new ListNode(1);
+        head.insertAtEnd(3);
+        head.insertAtEnd(4);
+        head.insertAtEnd(7);
+        head.insertAtEnd(1);
+        head.insertAtEnd(2);
+        head.insertAtEnd(6);
+
+        head.printList();
+        ListNode listNode = deleteMiddle(head);
+        listNode.printList(); // Output: [1,3,4,1,2,6]
+    }
+
+    public static ListNode deleteMiddle(ListNode head) {
+        // If the list has only one node, return null
+        if (head == null || head.next == null) {
+            return null;
+        }
+
+        ListNode slow = head;
+        ListNode fast = head;
+        ListNode prev = null;
+
+        // Move fast pointer two steps and slow pointer one step
+        while (fast != null && fast.next != null) {
+            fast = fast.next.next;
+            prev = slow;
+            slow = slow.next;
+        }
+
+        // Now 'slow' is pointing to the middle node
+        // 'prev' is pointing to the node before the middle node
+        // Remove the middle node by adjusting the 'next' pointer of the previous node
+        prev.next = slow.next;
+
+        return head;
+    }
+}