Remove Nth Node from End of List Leetcode

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.

The topic is to delete the node from the tail node n

The first is to find the node

Define two pointer p,q, two pointers to head node

The P-pointer goes first n steps, then two pointers go together, and the yourselves walk to the end of the hand, then the pointer after walking is just a distance from the tail node n

also save the previous node of the Q pointer, also note that if the deleted node is the head node , at which point the previous node of the Q node is not

The code is as follows:

<span style= "FONT-SIZE:18PX;" >/** * Definition for singly-linked list. * struct ListNode {*     int val; *     ListNode *next; *     listnode (int x): Val (x), Next (NULL) {} *}; */class Soluti On {public:    listnode *removenthfromend (listnode *head, int n) {                listnode *p,*q;//define P q,p go first n step, then p Q to traverse backwards at the same time, When P goes to the end, Q is at the bottom of the        p=head;         Q=head;                while (n>0)        {            p=p->next;            --n;        }         ListNode *pur=null;//is used to hold the previous node of the last nth node        while (P!=null)        {            pur=q;            q=q->next;            p=p->next;        }        if (pur==null)//indicates that the first node to be deleted is head change head        {            head=q->next;            Delete q;        }        else{                   pur->next=q->next;            Delete q;        }                    return head;                    }; </span>

Remove Nth Node from End of List Leetcode