Penultimate N nodes delete the list of Leetcod

Title Description

Given a list, delete list reciprocal of n nodes, and returns to the head node list.

Thinking

1. First traverse the list record length, before re-iterate to delete a position, delete skip bit. This method requires to traverse twice
2. Traversing double pointer, a pointer pointing to the start point A, when the pointer n + 1 step A proceeds, pointer B point to the starting point of beginning, when it reaches the end A, B happens to be a point before the deletion node

Code

method one:

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(head == NULL)
            return NULL;
        ListNode* cur = head;
        int count = 0;
        while(cur!= NULL)
        {
            count++;
            cur = cur->next;
        }
        if(n == count)
            return head->next;
        if(n == 0)
            return head;
        cur = head;
        int i = 0;
        while(i < count-n-1)
        {
            cur = cur->next;
            i++;
        }
        if(cur->next->next)
        {
            cur->next = cur->next->next;
        }
        else
        {
            cur ->next = NULL;
        }
        return head;
    }
};

Method Two:

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {   
        ListNode* dummy = new ListNode(NULL);
        dummy->next = head;  //添加头节点，便于操作
        ListNode* slow=dummy,* fast=dummy;
        int distance=0;
        while(fast->next){
            if(distance<n){
                fast=fast->next;
                distance++;
            }else{
                fast=fast->next;
                slow=slow->next;
            }
        }
        slow->next=slow->next->next;
        return dummy->next;
    }
};