比如删除的是头节点dummy=ListNode(0)dummy.next=head# 初始化快慢指针,初始时都指向哑节点slow=dummyfast=dummy# 快指针先前进n+1步,走到第n+1个节点# 这里加1是为了让快慢指针之间保持n的距离# 同时让慢指针停在要删除节点的前一个节点for_inrange(n+1):fast...
https://leetcode.com/problems/remove-nth-node-from-end-of-list/ 题目: nth For example, 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 do this in one...
1ListNode* removeNthFromEnd(ListNode* head,intn) {2ListNode** t1 = &head, *t2 =head;3//t2向后移n个节点4while(n--) t2 = t2->next;5//使t2移到最后一个节点的next,即NULL6//t1指向那个指向待删除节点的指针,即指向待删除节点的上一个节点的next7while(t2 !=NULL) {8t2 = t2->next;9...
Leetcode: Remove Nth Node From End of List 题目: 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->...
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 do this in one pass? 理解: 定义两个指针,一个在前,一个在后,同时遍历ListNode,要求是:开始遍历时fast指针要比slow指针快n步,方法是先让fast指针先移动...
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. 移除链表中倒数第n个节点; 写一个函数来确定要删除的是第几个节点,方法是获取链表总长度len,那么我们要删除的就是第 nth = len - n 个节点...
publicListNoderemoveNthFromEnd(ListNodehead,intn){intlen=0;ListNodeh=head;while(h!=null){h=h.next;len++;}//长度等于 1 ,再删除一个结点就为 null 了if(len==1){returnnull;}intrm_node_index=len-n;//如果删除的是头结点if(rm_node_index==0){returnhead.next;}//找到被删除结点的前一个结...
Given the head of a linked list, remove the nth node from the end of the list and return its head. Example 1:Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5] Example 2:Input: head = [1], n = 1 Out…
Given the head of a linked list, remove the nth node from the end of the list and return its head.impl Solution { pub fn remove_nth_from_end(head: Option<Box<ListNode>>, n: i32) -> Option<Box<ListNode>> { let mut dummy = Some(B
leetcode第19题 Given a linked list, remove the n-th node from the end of list and return its head. 给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。 题中的坑 这个题要注意的是:网站定义的链表结构head指向第一个有效元素,没有纯正意义上的头结点,我前两次提交就是因为这个问题没过...