237. Delete Node in a Linked List
by Botao Xiao
237. 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.
Given linked list – head = [4,5,1,9], which looks like following:
4 -> 5 -> 1 -> 9
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.
Thinking:
- Method 1:
- 没有给出链表头,我们将下一个值赋值到当前值。
- 将下一个结点删除。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void deleteNode(ListNode node) {
if(node == null) return;
node.val = node.next.val;
node.next = node.next.next;
}
}
二刷
- Same as what I did in previous time.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public void deleteNode(ListNode node) { ListNode cur = node; ListNode pre = null; while(cur.next != null){ cur.val = cur.next.val; pre = cur; cur = cur.next; } pre.next = null; } }
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