next return list ## 递归的解法 class Solution: def removeNthFromEnd(self, head, n): def remove(head): if not head: return 0, head i, head.next = remove(head.next) return i+1, (head, head.next)[i+1 == n] return remove(head)[1] 时间复杂度 这个算法的时间复杂度是 O(L),...
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...
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 yo...
public ListNode removeNthFromEnd(ListNode head, int n) { // nth指向p前n个结点,pre为nth前一个结点,即pre/nth/p ListNode p = head, nth = head, pre = null; while (--n > 0) { p = p.next; } while (p.next != null) { p = p.next; pre = nth; nth = nth.next; } // 此...
leetcode 19. Remove Nth Node From End of List 双指针 + 移动窗口,Givenalinkedlist,removethenthnodefromtheendoflistandreturnitshead.Forexample,Givenlinkedlist:1->2->3->4->5,andn=2.Afterremovingthesecondnodefromtheend,thel
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
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...
*/class Solution{public:ListNode*removeNthFromEnd(ListNode*head,intn){if(!head||!head->next){returnnullptr;}intnumber=0;ListNode*current=head;while(current){if(current->next){number+=2;current=current->next->next;}else{number+=1;current=current->next;}}inttarget=number-n;if(target==0)...
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
原问题链接:https://leetcode.com/problems/remove-nth-node-from-end-of-list/ 问题分析 这个问题相对来说比较简单。因为要删除链表中从后往前的第n个元素,所以需要用一个元素first的引用先往前n步,然后再用一个引用second跟着这个元素一步步到链表的最后。既然要删除这个倒数第n的元素,可以在第二个元素后面跟一...