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.
给一个链表，删除倒数第n个数字，我们的策略是两个指针，从头开始，假设倒数第五个，先让一个指针走出去五个，然后两个同时启动，当第一个到了最末尾，第二个也就到了我们最终想要的位置了。
java代码：
public static ListNode removeNthFromEnd(ListNode head, int n) {
if (n < 1) return head;
int i = 0;
ListNode before = head;
while (i < n + 1 && before != null) {
before = before.next;
i++;
}
if(i == n+1){
ListNode after = head;
while(before!=null){
before = before.next;
after = after.next;
}
ListNode temp = after.next;
after.next = temp.next;
}else if (i == n){
ListNode tmep = head;
head = head.next;
}
return head;
}
