Easy
Write a function to delete a node in a singly-linked list. You will not be given access to the head
of the list, instead you will be given access to the node to be deleted directly.
It is guaranteed that the node to be deleted is not a tail node in the list.
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.
Example 3:
Input: head = [1,2,3,4], node = 3
Output: [1,2,4]
Example 4:
Input: head = [0,1], node = 0
Output: [1]
Example 5:
Input: head = [-3,5,-99], node = -3
Output: [5,-99]
Constraints:
[2, 1000]
.-1000 <= Node.val <= 1000
node
to be deleted is in the list and is not a tail nodeimport com_github_leetcode.ListNode;
/*
* 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) {
while (node.next.next != null) {
node.val = node.next.val;
node = node.next;
}
node.val = node.next.val;
node.next = null;
}
}