Given a linked list, remove the nth node from the end of list and return its head.
Note: The solution set must not contain duplicate quadruplets.
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.
- 题目大意
给定一个单向链表，移除单向链表的倒数第n个。
比较简单，我的做法是将单向链表变成双向链表，找到最后一个，然后倒着找到倒数第n个节点并删除。 注意一些边界情况。
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
const removeNthFromEnd = function (head, n) {
let pointer = head;
while (pointer.next) {
pointer.next.last = pointer; //遍历，使其变为双向链表
pointer = pointer.next;
}
let i = 1;
while (i < n) {
pointer = pointer.last; //找到倒数第n个节点
i++;
}
if (pointer === head) { // 如果是head，直接将head向后移一个指针
head = head.next;
} else {
pointer.last.next = pointer.next;
}
return head;
};
