LeetCode: Remove Nth Node From End of List

Given 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
ListNode * p1 = head;
ListNode * p2 = head;
ListNode * pre = p2;
int count = 0;
while(count < n && p1){
count++;
p1 = p1->next;
}
if(count < n){
return NULL;
}
while(p1){
p1 = p1->next;
pre = p2;
p2 = p2->next;
}
if(pre == p2){
head = head->next;
}
else
pre->next = pre->next->next;
return head;
}
};