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:
题解输入:head = [1], n = 1
输出:[]
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
//快慢指针
ListNode dummy = new ListNode(0);
//虚拟头节点指向头节点
dummy.next = head;
//两个指针指向虚拟头节点
ListNode slow = dummy;
ListNode fast = dummy;
//让快指针走n步
while(n-- > 0){
fast = fast.next;
}
//然后一起走到最后
while(fast.next != null){
slow = slow.next;
fast = fast.next;
}
//此时的slow落在删除节点的前一个节点上
//删除节点
slow.next = slow.next.next;
return dummy.next;
}
}


![删除节点的倒数第N个节点[迭代]Day-10 删除节点的倒数第N个节点[迭代]Day-10](http://www.mshxw.com/aiimages/31/942275.png)
