給你一個(gè)鏈表，刪除鏈表的倒數(shù)第?n?個(gè)結(jié)點(diǎn)，并且返回鏈表的頭結(jié)點(diǎn)。
進(jìn)階：你能嘗試使用一趟掃描實(shí)現(xiàn)嗎？
示例 1：
輸入：head = [1,2,3,4,5], n = 2
輸出：[1,2,3,5]
示例 2：
輸入：head = [1], n = 1
輸出：[]
示例 3：
輸入：head = [1,2], n = 1
輸出：[1]
提示：
鏈表中結(jié)點(diǎn)的數(shù)目為 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
來(lái)源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list
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代碼實(shí)現(xiàn)：
class?Solution?{
????public?ListNode?removeNthFromEnd(ListNode?head,?int?n)?{
????????if?(head?==?null)?{
????????????return?head;
????????}
????????List<ListNode>?list?=?new?ArrayList<>();
????????ListNode?it?=?head;
????????while(it?!=?null)?{
????????????list.add(it);
????????????it?=?it.next;
????????}
????????int?num?=?list.size();
????????if?(n?==?num)?{
????????????if?(num?==?1)?{
????????????????return?null;
????????????}?else?{
????????????????head?=?list.get(1);
????????????}????????????
????????}?else?if?(n?>?num)?{
????????????return?head;
????????}?else?if?(n?==?1)?{
????????????list.get(num?-?n?-?1).next?=?null;
????????}?else?{
????????????list.get(num?-?n?-?1).next?=?list.get(num?-?n?+?1);
????????}
????????return?head;
????}
}
