"Leetcode" 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.

Answer

Using the idea of quick and slow pointers, the fast hands go first n steps, then go at the same time and stop at the end of the pointer. Note that the processing of the head node, such as 1->2->null, n=2, returns 2->NULL, the code is as follows:

/** * Definition for singly-linked list. * public class ListNode {*     int val; *     ListNode Next; *     listnode (int x) {*         val = x; *         next = null; *< c5/>} *} */public class Solution {public    ListNode removenthfromend (listnode head, int n) {        if (head==null) {            R Eturn head;        }        ListNode Fast=head;        ListNode Slow=head;        while (n-->0) {            fast=fast.next;        }        if (fast==null) {        //Determine if fast is at the end of the            return slow.next;        }        while (fast.next!=null) {//note here, fast must not be null, so the previous case of NULL is            Fast=fast.next;            Slow=slow.next;        }        Slow.next=slow.next.next;        return head;    }}

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Leetcode Remove Nth Node from End of List