Remove Nth Node From End of List @ LeetCode

Package Level2; import Utility. listNode;/***** Remove Nth Node From End of List ** Given a linked list, remove the nth node from the end of list and return its head. for example, Given linked list: 1-> 2-> 3-> 4-> 5, and n = 2. after removing the second node from the end, the linked list becomes 1-> 2-> 3-> 5. note: Given n will always be valid. try to do this in one pass. */public class S19 {public Static void main (String [] args) {ListNode head = new ListNode (1); ListNode l2 = new ListNode (2); ListNode l3 = new ListNode (3 ); listNode l4 = new ListNode (4); ListNode l5 = new ListNode (5); head. next = l2; l2.next = l3; l3.next = l4; l4.next = l5; ListNode h = removeNthFromEnd (head, 5); h. print ();} public static ListNode removeNthFromEnd (ListNode head, int n) {if (n = 0 | head = null) {return Head;} if (n = 1 & head. next = null) {return null;} ListNode p = head, q = head; // Let p first q n positions for (int I = 0; I <n; I ++) {if (p! = Null) {p = p. next;} else {return head ;}// if p is already null at this time, it indicates that the deleted header must be if (p = null) {head = head. next; return head;} // p and q move forward together while (p. next! = Null) {q = q. next; p = p. next;} // Delete the element q. next = q. next. next; return head ;}}