leetcode: Remove Nth Node From End of List

iven a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Try to do this in one pass.

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {

if (!head)
return NULL;

ListNode *pDistantNode = head;
ListNode *pPreNode;
int count = n-1;

while (pDistantNode && count > 0)
{
pDistantNode = pDistantNode->next;
count--;
}

if (pDistantNode->next == NULL)
return head->next;

ListNode *pCurNode = head;
while (pDistantNode->next)
{
if (pDistantNode->next->next == NULL)
{
pPreNode = pCurNode;
}

pDistantNode = pDistantNode->next;
pCurNode = pCurNode->next;
}

pPreNode->next = pCurNode->next;

return head;
}
};