[Leetcode]Remove Nth Node From End of List

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.

思路是用一个fast指针和一个slow指针，fast先跑，比slow多跑n的点。然后fast和slow一起开始跑，当fast跑到最后一个点的时候，slow.next正好是倒数第n个点，需要删除。实现过程中注意使用一个头部哑节点来考虑corner case。

public ListNode removeNthFromEnd(ListNode head, int n) {
if (head == null)
return head;
ListNode dump = new ListNode(0);
dump.next = head;
ListNode fast = dump;
ListNode slow = dump;

for (int i = 0; i < n; i++) {
fast = fast.next;

if (fast == null)
break;
}

while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return dump.next;
}