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;
        }
    }
}
原文地址:https://www.cnblogs.com/wxisme/p/4687486.html