原题链接:https://leetcode.com/problems/delete-node-in-a-linked-list/ 题目描述: Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. 1 -> 2 -> 3 -> 4and you are given the third node with value3, the linked list should ...
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is1 -> 2 -> 3 -> 4and you are given the third node with value3, the linked list should become1 -> 2 -> 4after calling your function. 分析:...
node的val用node.next.val代替,然后node.next指向node.next.next,这样即使是单向链表,删除节点还是不会断。 代码实现: #Definition for singly-linked list.#class ListNode(object):#def __init__(self, x):#self.val = x#self.next = NoneclassSolution(object):defdeleteNode(self, node):""":type nod...
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Given linked list – head = [4,5,1,9], which looks like following: 4 -> 5 -> 1 -> 9 1. Example 1: Input: head = [4,5,1,9], node = 5 Output: [4,1,9] ...
*/public class Solution{publicvoiddeleteNode(ListNode node){*node=*node->next;}} 疑问: 用Java像C++那样,直接将 node = node.next 将不能ac,必须对其val和next分别设置,这是为什么呢?希望高手帮忙解答一下~ 合集:https://github.com/Cloudox/LeetCode-Record...
public TreeNode deleteNode(TreeNode root, int key) { if (root == null) { return null; } if (root.val > key) { root.left = deleteNode(root.left, key); } else if (root.val < key) { root.right = deleteNode(root.right, key); ...
image.png 方法: 这题最难的是理解题意,输入不是链表的头,而是要删除的结点,所以只需要将该结点的值改为下一个结点的值,同时把它的下一个指向改为下下个结点。 classDeleteNodeInALinkedList{classListNode(var`val`:Int){varnext:ListNode?=null}fundeleteNode(node:ListNode?){node?.`val`=node?.next?....
力扣leetcode-cn.com/problems/delete-node-in-a-linked-list/ 题目描述 请编写一个函数,使其可以删除某个链表中给定的(非末尾)节点,你将只被给定要求被删除的节点。 现有一个链表 -- head = [4,5,1,9],它可以表示为: 示例1: 输入: head = [4,5,1,9], node = 5 输出: [4,1,9] 解...
LeetCode-237-删除链表中的节点 删除链表中的节点 题目描述：请编写一个函数，使其可以删除某个链表中给定的（非末尾）节点。传入函数的唯一参数为 要被删除的节点。示例说明请见LeetCode官网。来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/delete-node-in-a-linked-list/著作权归领扣网络所有。
TreeNode* deleteNode(TreeNode* root, int key) 确定终止条件 遇到空返回,其实这也说明没找到删除的节点,遍历到空节点直接返回了 if (root == nullptr) return root; 确定单层递归的逻辑 这里就把平衡二叉树中删除节点遇到的情况都搞清楚。 有以下五种情况: 第一种情况:没找到删除的节点,遍历到空节点直接...