Given a linked list, remove then-th node from the end of list and return its head. Example: Given linked list:1->2->3->4->5, andn= 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Givennwill always be valid. Follow up: Could you...
Method 1:先遍历一次获得中长度, 在遍历len-n+1次获得所求node Method 2:运用双指针, 开始两个指针都指向head, 然后先移动第一个指针, 使其指向第n个node, 然后两个指针一个移动, 当第一个指针移动到尾部时, 第二个node就指向所求node code 2015-08-16...
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 d...
https://leetcode.com/problems/remove-nth-node-from-end-of-list/ 题目: nth For example, AI检测代码解析 Given linkedlist:1->2->3->4->5,andn=2.After removing the second node from the end,the linkedlistbecomes1->2->3->5. 1. 2. 3. Note: Given n will always be valid. Try to d...
After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Given n will always be valid. Follow up: Could you do this in one pass? 思路: 这道题让我们移除链表倒数第N个节点,限定n一定是有效的,即n不会大于链表中的元素总数。还有题目要求我们一次遍历解...
19. 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, After removing the second node from the end, the linked list bec...
Can you solve this real interview question? Remove Nth Node From End of List - Given the head of a linked list, remove the nth node from the end of the list and return its head. Example 1: [https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.j
19. Remove Nth Node From End of List 题目描述 题目解读 一开始做的时候以为一直是倒数第二个,我想这不是很容易,结果没看清题目,那个bug改的痛啊,以及要纪念一下第一次不看讨论区独立完成的代码。 代码 Leetcode 19. Remove Nth Node From End of List 题目描述:删除从尾节点数其的第n个节点,并返回删...
LinkedList的removeNthFromEnd方法的时间复杂度是多少? 19. Remove Nth Node From End of List Given a linked list, remove the n-th node from the end of list and return its head. Example: 代码语言:javascript 代码运行次数:0 运行 AI代码解释Given...
(head.next == null) return null; //list只有一个node,潜台词就是n=1 ListNode newhead = new ListNode(Integer.MIN_VALUE); //被删除的节点可能是head,故设置一个newhead newhead.next = head; ListNode p = head; //利用双引用实现一次遍历即删除倒数节点的目的 ListNode q = head; //示例list:1...