[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.

去除某个节点，需要维护改节点前面的节点，因为是单向链表，使用两个指针，其中一个比另一个快n步，这里由给出的n都是合法的，因此不需要特殊考虑，不过在实际中若需要考虑n为不合法的情况，则需要直接返回。

public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null || n == 0) return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode temp1 = head;
        while(n > 0) {
            temp1 = temp1.next;
            n--;
        } // since n is valid, there is no need to check here, but should discuss with interviewer. If n is invalid, terminate here
        ListNode temp = dummy, temp2 = head;
        while(temp1 != null) {
            temp1 = temp1.next;
            temp2 = temp2.next;
            temp = temp.next;
        }
        temp.next = temp2.next;
        
        return dummy.next;
    }