Given a linked list, remove the n-th node from the end of list and return its head.
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.
题目
删除链表倒数第n个元素
思路
经典题目
快慢指针法，快指针先走n步，再和慢指针一起走，当快指针走到最后一个元素时，慢指针指向倒数第n个元素。
这里有一个技巧：就是建立虚拟头结点，快慢指针从虚拟头结点开始遍历，这样最后慢指针会指向倒数第n个元素的前驱，传统方法删除即可。
若从head开始遍历，则需要考虑n=1的情况，也就是慢指针指向最后一个节点并删除。
代码
public ListNode removeNthFromEnd(ListNode head, int n) {
if (head.next == null)
return null;
ListNode dummy = new ListNode(0);
ListNode fast = dummy, slow = dummy;
dummy.next = head;
for (int i = 0; i < n; ++i)
fast = fast.next;
while (fast.next != null) {
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return dummy.next;
}
