Leetcode 19. Remove Nth Node from End of List Java language

Given A linked list, remove the nth node from the end of the 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'll always be valid . Try to do the in one pass.

Test instructions: Deletes the nth node of the bottom. Traverse as much as possible.

/** * definition for singly-linked list. * public class listnode  { *     int val; *     ListNode  Next; *     listnode (int x)  { val = x; } *  } */public class Solution {    public ListNode  Removenthfromend (listnode head, int n)  {        if (head==null | | &NBSP;N&LT;1) return head;        listnode cur=head;         while (cur!=null) {             n--;            cur= Cur.next;        }        if (n ==0) head=head.neXt;        if (n<0) {             cur=head;            while (++n!=0) {                cur=cur.next;             }             cur.next=cur.next.next;        }         return head;             }}

PS: The left teacher provides a way to find the countdown to the first K. Go through it, k--at the same time, and finally judge the size of K. K==0 indicates that the head node is to be deleted. K>0, stating that K is wrong. K<0 time, and then go from the beginning again, but at this time K++,k=0 also found the first K-1. Simply delete the line at this time.

"Method 2" online fast and fast pointer method. Fast goes K-1 step first. Then slow and fast step-by-step. Fast arrives at the end of the slow to the penultimate K-1.

/*public class listnode {    int val;    listnode  next = null;     listnode (Int val)  {         this.val = val;    }}*/public class  Solution {    public listnode findkthtotail (ListNode head,int k)  {        if (head==null | |  k==0) return null;        listnode fast=head;         for (int i=0;i<k-1;i++) {             ////Prevent K Invalid              if (fast.next!=null) {                 fast=fast.next;               }else{                 return null;             }                      }         Listnode slow=head;        while (fast.next!=null) {             slow=slow.next;             fast=fast.next;        }         return slow;     }}

Leetcode 19. Remove Nth Node from End of List Java language