[Java]LeetCode237. 刪除鏈表中的節點 | Delete Node in a Linked List

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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:

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

Note:

• The linked list will have at least two elements.
• All of the nodes' values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

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請編寫一個函數，使其能夠刪除某個鏈表中給定的（非末尾）節點，你將只被給定要求被刪除的節點。

現有一個鏈表 -- head = [4,5,1,9]，它能夠表示爲:

    4 -> 5 -> 1 -> 9

示例 1:

輸入: head = [4,5,1,9], node = 5
輸出: [4,1,9]
解釋: 給定你鏈表中值爲 5 的第二個節點，那麼在調用了你的函數以後，該鏈表應變爲 4 -> 1 -> 9.

示例 2:

輸入: head = [4,5,1,9], node = 1
輸出: [4,5,9]
解釋: 給定你鏈表中值爲 1 的第三個節點，那麼在調用了你的函數以後，該鏈表應變爲 4 -> 5 -> 9.

說明:

• 鏈表至少包含兩個節點。
• 鏈表中全部節點的值都是惟一的。
• 給定的節點爲非末尾節點而且必定是鏈表中的一個有效節點。
• 不要從你的函數中返回任何結果。

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0ms

1 class Solution {
2     public void deleteNode(ListNode node) {
3         node.val=node.next.val;
4         node.next=node.next.next;
5     }
6 }

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34180 kb

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public void deleteNode(ListNode node) {
11         
12         int temp;
13         temp = node.val;
14         node.val = node.next.val;
15         node.next.val =  temp;
16         
17         node.next = node.next.next;        
18         
19     }
20 }