题目：
给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。
示例 1：
输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]
示例 2：
输入：head = [1], n = 1
输出：[]
示例 3：
输入：head = [1,2], n = 1
输出：[1]
提示：
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
解决方法：
1.快慢指针
fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
var dump = ListNode(-1)
dump.next = head
var slow = dump
var fast = head
for (index in 1..n) {
fast = fast?.next
}
while (fast != null) {
fast = fast.next
slow = slow.next
}
slow.next = slow.next.next
return dump.next
}
2.递归
var curIndex = -1
fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
var dump:ListNode = ListNode(-1,head)
removeReverse(dump,n)
return dump.next
}
fun removeReverse(head: ListNode?, n: Int) {
if (head == null){
curIndex = 0
return
}
removeReverse(head.next,n)
if (curIndex >= 0){
if (++curIndex == n+1){
head.next = head.next.next
}
}
}
3.使用栈：
fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
var stack = LinkedList<ListNode>()
var dump :ListNode? = ListNode(-1,head)
var cur = dump
while (cur != null){
stack.push(cur)
cur = cur.next
}
for (i in 1..n){
stack.pop()
}
val pop = stack.pop()
pop.next = pop.next.next
return dump?.next
}
