删除单向链表倒数第n个节点

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) {
ListNode *p = head;
while (n > 0 && p != NULL)
{
p = p->next;
n--;
}

if (n != 0)
{
return head;
}

if (n == 0 && p == NULL)
{
ListNode *temp = head->next;
delete head;
return temp;
}

ListNode *q = head;
while (p->next != NULL)
{
p = p->next;
q = q->next;
}

ListNode *temp = q->next->next;
delete q->next;
q->next = temp;

return head;
}
};