LeetCode -- Delete Node in a Linked List

Question: 

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.

Analysis: 

题目只给出要删除的节点外没有给出任何其他信息。但是在单链表中我们知道,要删除一个节点,必须给出上一个节点。在这里,题目只给出了目前节点,因此我们可以删除的只有它后面的节点。因此思路就是:想办法把后面节点变为我们要删除的节点。用后面一个节点的值覆盖当前节点的值,这样后面的节点就成来没用的废气节点,然后删除即可。

Answer:

/**
 * 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) {
         if(node.next.next == null){
                node.val = node.next.val;
                node.next = null;
        }
        else {
                node.val = node.next.val;
                node.next = node.next.next;
        }
    }
}
原文地址:https://www.cnblogs.com/little-YTMM/p/4784956.html