Lettcode_237_Delete Node in a Linked List

本文是在学习中的总结,欢迎转载但请注明出处:http://blog.csdn.net/pistolove/article/details/47334649

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.

思路:

(1)该题为给定一个链表中的一个节点,要求删除该节点。

(2)该题主要对链表存储的考察。在一般情况下,要删除链表的一个节点,需要知道当前节点的前驱节点。该题没有给出前驱节点,所以就需要将当前节点的后继节点的值复制到当前节点,然后删除后继节点即可。

(3)该题比较简单。详情见下方代码。希望本文对你有所帮助。

算法代码实现如下:

package leetcode;

import leetcode.utils.ListNode;

/**
 *
 * @author lqq
 *
 */
public class Delete_Node_in_a_LinkedList {

	public void deleteNode(ListNode node) {
		if (node == null)
			return;

		ListNode after = node.next;
		if (after != null) {
			int value = after.val;
			node.val = value;
			node.next = after.next;
		}
	}
}

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时间: 2024-10-12 12:34:15

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