解题思路：

链表题目：哑节点、栈、快慢指针（双指针）

方法一：计算链表长度

java">class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode dum = new ListNode(0, head);int len = getLen(head);ListNode cur = dum;for (int i = 0; i < len - n; i++) {cur = cur.next;}cur.next = cur.next.next;return dum.next;}public int getLen(ListNode head) {int len = 0;while (head != null) {head = head.next;len++;}return len;}
}

方法二：栈

java">class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode dum = new ListNode(0, head);Deque<ListNode> stack = new LinkedList<>();ListNode cur=dum;while(cur!=null){stack.push(cur);cur=cur.next;}for(int i=0;i<n;i++){stack.pop();}ListNode pre=stack.peek();pre.next=pre.next.next;return dum.next;}
}

方法三：快慢指针（双指针）

注意fast从head开始，而不是dum

java">class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode dum = new ListNode(0, head);ListNode fast=head;ListNode slow=dum;for(int i=0;i<n;i++){fast=fast.next;}while(fast!=null){slow=slow.next;fast=fast.next;}slow.next=slow.next.next;return dum.next;}
}