Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

考虑用双链表，先让第一个链表走n-1步，然后让第二个链表从头开始两个链表一起走，直到第一个走到头，此时第二个链表刚好走到要删除的节点，保存删除节点的上一个节点即可，但是要考虑边界条件，这里考虑了n输入不合法的问题

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *removeNthFromEnd(struct ListNode *head, int n) {
    if(head == NULL || n<= 0) return NULL;
    struct ListNode *pre = NULL;
    struct ListNode *p = head;
    struct ListNode *q = head;
    n--;
    while(p->next != NULL && n>0){
        p = p->next;
        n--;
    }
    if(n>0) return NULL;
    while(p->next != NULL){
        pre = q;
        p = p->next;
        q = q->next;
    }
    if(pre == NULL){
        head = q->next;
    }else{
        pre->next = q->next;
    }
    return head;
}

看到还有一种方法是走到要删除的节点时，将它的下一个节点值复制到当前节点，然后删除它的下一个节点，这种方法也可以。但是要判断倒数第一个节点的问题。