LinkedLists.DeleteNode

basic approach is to change connection node.next = node.next.next

1. RemoveElements(ListNode head, int val)

leetcode task

optimal solution through sentinel or fake node

/**
 * 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; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        //degenerate case
        if (head == null) return null;
        
        //creating additional node to the left from first and a variable "sentinel" with reference to it
        ListNode node;
        ListNode sentinel = new ListNode();
        sentinel.next = head;
        node = sentinel;
        
        //checking node value and deleting reference to it if conditions are met
        while (node.next.next != null){
            if (node.next.val == val){
                node.next = node.next.next;
                continue;
            }         
             node = node.next;
        }
        
        //edge case (last ListNode element)
       if (node.next.val == val) node.next = null;
        
       return sentinel.next;
    }
}

not optimal solution with border cases

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode RemoveElements(ListNode head, int val) {
        
        if (head == null) 
            return null;
        
        ListNode res = head;
        
        ListNode current = head;
        ListNode previous = null;
        // bool wasDeletedFromMiddle = false;
        
        while (current != null) {
            
            if (current.val == val) {
                
              if (current == head) {
                head = head.next;                 
                res = head;
             }
              else if (current != head && current.next != null) {
                  if (current.next != null && current.next.val == val) {
                      current = current.next;
                      continue;
                  }
                    
                  previous.next = current.next;
                  
              }
             // tail
             else if (current != head && current.next == null) {
                 previous.next = null;
             }                      
                
            }   
            
                        
            previous = current;            
            current = current.next;
        }

        return res;
   
    }
}

2. DeleteNode(ListNode node)

leetcode

Here we don’t have access to previos node

    public void DeleteNode(ListNode node) {
      node.val = node.next.val;
      node.next = node.next.next;  
    }