删除单链表的倒数第n个元素

题目原型：

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.

这个题目比较简单，就不分析了，直接贴代码。

public ListNode removeNthFromEnd(ListNode head, int n)
{
//求出链表长度
int len = 0;//链表长度
ListNode p,q;
p = head;
while(p!=null)
{
len++;
p = p.next;
}
if(n>len)
return null;

p = head;
q = head;
for(int i = 0;i<len-n;i++)
{
p = q;
q = q.next;
}
//当删除的是第一个节点的时候
if(p==q)
head = p.next;
if(q!=null)
{
p.next = q.next;
q.next = null;
}

return head;
}