"Leetcode" 19. Remove Nth Node from End of List

Topic:

Idea: If the list is empty or n is less than 1, return directly, otherwise, let the list go from head to tail, each move one step, let N minus 1.

1. List 1->2->3,n=4, there is no fourth node, return the entire list

Swept nodes in sequence:

n Worth changing: 3-2-1

2. Linked list 1->2->3,n=3

Swept nodes in sequence:

n Worth changing: 2-1-0

3. Linked list 1->2->3,n=2

　　　　　　Swept nodes in sequence:

　　　　　　n Worth changing: 1-0--1

When walking to the end of the list: 1.n>0, indicating that the list does not have the nth node, directly return to the original linked list;

2.n=0, indicating that the last nth node of the list is the head node, returning to the Head.next;

3.n<0, start from the beginning, do not move one step to let N plus 1, when the n=0, move stop, the current move to the node is to delete the node's previous node. Because if the list length is L, then the previous node of the nth-L-n is the second. When the first sweep to the end of the list, the value of n becomes n-l, and when N is added 1 until 0 o'clock, the second sweep is exactly the first l-n node.

/** * Definition for singly-linked list. * public class ListNode {*     int val; *     ListNode Next; *     listnode (int x) {val = x;}}} */public class Soluti On {public    ListNode removenthfromend (listnode head, int n) {        if (head==null| | n<1) {            return head;        }         ListNode Cur=head;         while (cur!=null) {             n--;             Cur=cur.next;         }         if (n==0) {             return head.next;         }         if (n<0) {             cur=head;             while (++n!=0) {//When n=0 moves stop, the node moved to is the previous node to delete the node                 cur=cur.next;             }             Cur.next=cur.next.next;         }         return head;    }}

　　

"Leetcode" 19. Remove Nth Node from End of List