LeetCode(Remove Nth Node From End of List)删除链表倒数第n个节点

题目要求:

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.
代码：

class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
if(head == NULL)
return NULL;
if(n <= 0)
return head;
ListNode* first_ptr = head, *second_ptr = head;
for(size_t i = 0; i < n - 1; ++i)
{
if(first_ptr->next != NULL)
first_ptr = first_ptr->next;
else
return head;
}
if(first_ptr->next == NULL)//如果要删除的是头结点
{
ListNode* node = head;
head = head->next;
delete node;
node = NULL;
return head;
}
ListNode* pre = head;//保存要删除节点的前一个节点
if(first_ptr->next != NULL)
{
first_ptr = first_ptr->next;
second_ptr = second_ptr->next;
}
while(first_ptr->next != NULL)
{
first_ptr = first_ptr->next;
second_ptr = second_ptr->next;
pre = pre->next;
}
pre->next = second_ptr->next;
delete second_ptr;
second_ptr = NULL;
return head;
}
};