Leetcode | #19 Remove Nth Node from End of List

Topic:

Given A linked list, remove the nth node from the end of the list and return its head.

For example,

   1->2->3->4->5  N = 2.   1->2->3->5.

Note:
Given n would always be valid.
Try to do the in one pass.

Ideas:

• Count the number of nodes first, and change the number of count-n number of next to count-n-2.
• The topic requires the best to traverse, that is mathematically clever, two pointers, first to go N steps, and then start and second at the same time, until the next is empty, then the second next to next
  

public class Removenthfromend {public class ListNode {int val; ListNode Next; ListNode (int x) {val = x;    next = null;}} /*//first count the number of nodes count, the number of count-n number of the next change to the number of count-n-2 public ListNode removenthfromend (listnode head, int n) {int count = 1; ListNode temp = Head;while (Temp.next! = null) {count++;temp = Temp.next;} int i = count-n;if (i = = 0) {return head.next;} temp = Head;while (i > 1) {temp = temp.next;i--;} Temp.next = Temp.next.next;return head;    } *///the topic requires the best to traverse, that is, mathematically clever, two pointers, first to go n steps,//and second at the same time walk, until the first next is empty, the second next change to next Nextpublic ListNode Removenthfromend (listnode head, int n) {ListNode first = head, second = head;for (int i=n; i>0; i--) {first = he Ad.next;} if (first = = null) {return first;} while (first.next! = null) {first = First.next;second = Second.next;} Second.next = Second.next.next;return head;    }}

Leetcode | #19 Remove Nth Node from End of List