删除链表 M 个节点后的 N 个节点的 Javascript 程序
原文:https://www . geesforgeks . org/JavaScript-program-to-delete-n-nodes-after-m-nodes-of-a-link-list/
给定一个链表和两个整数 M 和 N。遍历链表，保留 M 个节点，然后删除下 N 个节点，一直到链表结束。 难度等级:菜鸟 示例:
Input:
M = 2, N = 2
Linked List: 1->2->3->4->5->6->7->8
Output:
Linked List: 1->2->5->6
Input:
M = 3, N = 2
Linked List: 1->2->3->4->5->6->7->8->9->10
Output:
Linked List: 1->2->3->6->7->8
Input:
M = 1, N = 1
Linked List: 1->2->3->4->5->6->7->8->9->10
Output:
Linked List: 1->3->5->7->9
问题的主要部分是维护节点之间的适当链接，确保所有的角落案例都得到处理。下面是 skipMdeleteN()函数的 C 实现，它跳过 M 个节点，删除 N 个节点，直到列表结束。假设 M 不能为 0。
java 描述语言
<script>
// Javascript program to delete N nodes
// after M nodes of a linked list
// A linked list node
class Node
{
constructor()
{
this.data = 0;
this.next = null;
}
}
// Function to insert a node at
// the beginning
function push(head_ref, new_data)
{
// Allocate node
var new_node = new Node();
// Put in the data
new_node.data = new_data;
// Link the old list off the
// new node
new_node.next = (head_ref);
// Move the head to point to
// the new node
(head_ref) = new_node;
return head_ref;
}
// Function to print linked list
function printList(head)
{
var temp = head;
while (temp != null)
{
document.write(temp.data + " ");
temp = temp.next;
}
document.write("<br/>");
}
// Function to skip M nodes and then
// delete N nodes of the linked list.
function skipMdeleteN(head, M , N)
{
var curr = head, t;
var count;
// The main loop that traverses
// through the whole list
while (curr != null)
{
// Skip M nodes
for (count = 1; count < M &&
curr != null; count++)
curr = curr.next;
// If we reached end of list,
// then return
if (curr == null)
return;
// Start from next node and delete
// N nodes
t = curr.next;
for (count = 1; count <= N &&
t != null; count++)
{
var temp = t;
t.next;
}
// Link the previous list with
// remaining nodes
curr.next = t;
// Set current pointer for next
// iteration
curr = t;
}
}
// Driver code
/* Create following linked list
1.2.3.4.5.6.7.8.9.10 */
var head = null;
var M = 2, N = 3;
head = push(head, 10);
head = push(head, 9);
head = push(head, 8);
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
document.write(
"M = "+M+", N = " + N +
"<br/>Given Linked list is :<br/>");
printList(head);
skipMdeleteN(head, M, N);
document.write(
"<br/>Linked list after deletion is :<br/>");
printList(head);
// This code is contributed by gauravrajput1
</script>
输出:
M = 2, N = 3
Given Linked list is :
1 2 3 4 5 6 7 8 9 10
Linked list after deletion is :
1 2 6 7
时间复杂度: O(n)，其中 n 为链表中的节点数。
更多详情请参考完整文章删除链表M 个节点后的 N 个节点！
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