The Fellow Programmer

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Given linked list -- head = [4,5,1,9], which looks like following: Example 1: Input: head = [4,5,1,9], node = 5 Output: [4,1,9] Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function. Example 2: Input: head = [4,5,1,9], node = 1 Output: [4,5,9] Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function. Note: The linked list will have at least two elements. All of the nodes' values will be unique. The given node will not be the tail and it will always be a valid node of the linked list.   /**  * Definition for singly-linked list.  * public class ListNode {  *     int val;  *     ListNode next;  *     ListNode(int x) { val = x; }  * }  */ class Solution {     public void deleteNode(ListNode node) {  

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity. Example 1: Input: 1->2->3->4->5->NULL Output: 1->3->5->2->4->NULL Example 2: Input: 2 ->1->3->5->6->4->7->NULL Output: 2->3->6->7->1->5->4->NULL Note: The relative order inside both the even and odd groups should remain as it was in the input. The first node is considered odd, the second node even and so on ...   /**  * Definition for singly-linked list.  * public class ListNode {  *     int val;  *     ListNode next;  *     ListNode() {}  *     ListNode(int val) { this.val = val; }  *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }  * }  */ public class Solution {     public List

Given a non-empty, singly linked list with head node head , return a middle node of linked list. If there are two middle nodes, return the second middle node. Example 1: Input: [1,2,3,4,5] Output: Node 3 from this list (Serialization: [3,4,5] ) The returned node has value 3. (The judge's serialization of this node is [3,4,5]). Note that we returned a ListNode object ans, such that: ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.   Example 2: Input: [1,2,3,4,5,6] Output: Node 4 from this list (Serialization: [4,5,6] ) Since the list has two middle nodes with values 3 and 4, we return the second one. Here it can be done in two ways. Option 1: 1)Take two pointers one as slow and other as fast. 2)Initially both pointing to the head. 3)Slow points the next element and fast points next of next element. 4)Step 3 has to repeated until fast equal to null and fast of next equal to null. 5)After this point, slow will poin