Leetcode 237 Delete Node In A Linked List

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:

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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:

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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:

1 2	Input: head = [1,2,3,4], node = 3 Output: [1,2,4]

Example 4:

1 2	Input: head = [0,1], node = 0 Output: [1]

Example 5:

1 2	Input: head = [-3,5,-99], node = -3 Output: [5,-99]

Constraints:

The number of the nodes in the given list is in the range [2, 1000].
-1000 <= Node.val <= 1000
The value of each node in the list is unique.
The node to be deleted is in the list and is not a tail node

交换节点的值

一开始想这个题总想不出来, 感觉必须要获得这个节点前面的一个节点才能删除.

后来看了答案才发现只需要交换值就可以!!!

把这个节点的后一个节点的值覆盖到这个节点上, 然后删除后一个节点就可以了!!

可能是之前做链表题, 有很多链表题都要求只能进行指针的计算不能交换值这个条件限制了思路.

还是应该打开思路, 多多创新.

/**
 * 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.next.next == null)
        {
            node.val = node.next.val;
            node.next = null;
        }
        else
        {
            node.val = node.next.val;
            node.next = node.next.next;
        }
    }
}