def removeNthFromEnd(self, head, n): 类的一个方法,用于删除链表的倒数第n个节点。 def remove(head): 一个内部定义的递归函数,用来递归地遍历链表,同时找到并删除指定的节点。 递归函数 remove 这个递归函数是解决方案的核心。递归意味着函数会调用自身来解决问题的子部分,直到达到一个基本情况(base case),然...
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...
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. 思路分析: 双指针,前后两个指针之间的距离为n-1,当后一个指针指向链表末尾结点的时候,前一个指针指向的结点就是要删除的结点。
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...
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...
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每天一算:Remove Nth Node From End of List LeetCode上第19号问题:Remove Nth Node From End of List 题目 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。 示例: 给定一个链表: 1->2->3->4->5, 和 n = 2. 当删除了倒数第二个节点后,链表变为 1->2->3->5 ...
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 linked list: **1->2->3->4->5**, and **_n_ = 2**. After removing the second node from...