问题
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请编写一个函数，使其可以删除某个链表中给定的（非末尾）节点，你将只被给定要求被删除的节点。
现有一个链表 -- head = [4,5,1,9]，它可以表示为:
4 -> 5 -> 1 -> 9
输入: head = [4,5,1,9], node = 5
输出: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.
输入: head = [4,5,1,9], node = 1
输出: [4,5,9]
解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.
说明:
链表至少包含两个节点。
链表中所有节点的值都是唯一的。
给定的节点为非末尾节点并且一定是链表中的一个有效节点。
不要从你的函数中返回任何结果。
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
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.
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.
示例
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public class Program {
public static void Main(string[] args) {
var head = new ListNode(1) {
next = new ListNode(2) {
next = new ListNode(3) {
next = new ListNode(4) {
next = new ListNode(5)
}
}
}
};
DeleteNode(head.next.next);
ShowArray(head);
Console.ReadKey();
}
private static void ShowArray(ListNode list) {
var node = list;
while(node != null) {
Console.Write($"{node.val} ");
node = node.next;
}
Console.WriteLine();
}
private static void DeleteNode(ListNode node) {
//先复制下一个节点的值
node.val = node.next.val;
//再把下一节点的指针指向下下一个节点
node.next = node.next.next;
}
public class ListNode {
public int val;
public ListNode next;
public ListNode(int x) { val = x; }
}
}以上给出1种算法实现，以下是这个案例的输出结果：
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1 2 4 5分析：
显而易见，以上算法的时间复杂度为： 