LeetCode——Delete Node in a Linked List
Description:
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 is 1 -> 2 -> 3 -> 4
and you are given the third node with value 3
, the linked list should become 1 -> 2 -> 4
after calling your function.
删除单向链表的一个节点,只给出了要删除的节点。
思路:从要删除的节点的下一个节点开始,逐一覆盖前面的节点的值,注意要删除最后一个多余的节点(Java中指向空即可,等待垃圾回收)。
代码:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public void deleteNode(ListNode node) { ListNode iNode = node; while(iNode.next != null) { //覆盖当前节点的值 iNode.val = iNode.next.val; //删除最后一个多余的节点 if(iNode.next.next == null) { iNode.next = null; break; } iNode = iNode.next; } } }
作者:Pickle
声明:对于转载分享我是没有意见的,出于对博客园社区和作者的尊重一定要保留原文地址哈。
致读者:坚持写博客不容易,写高质量博客更难,我也在不断的学习和进步,希望和所有同路人一道用技术来改变生活。觉得有点用就点个赞哈。
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