LeetCode: Remove Nth Node From End of List

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.

思路： 先让一个指针走n步，在让另一个指针一起向前。需要注意n的取值。有可能直接删除头指针，需要判断

class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
if (0 == n)
return head;
int c = 0;
ListNode *first = head, *second=head;

while (c++ < n){
first = first->next;
}
if (first == NULL){
return head->next;
}
while ( first->next != NULL){
first = first->next;
second = second->next;
}
ListNode *p = second->next;
second->next = p->next;
delete p;
return head;

}
};