用于就地重新排列给定链表的 Javascript 程序
原文:https://www . geeksforgeeks . org/JavaScript-program-for-重排-给定-就地链表/
给定一个单链表 L0->L1->……->Ln-1->Ln。重新排列列表中的节点，使新形成的列表为:L0->Ln->L1->Ln-1->L2->Ln-2…… 您需要在不改变节点值的情况下就地执行此操作。
示例:
Input: 1 -> 2 -> 3 -> 4
Output: 1 -> 4 -> 2 -> 3
Input: 1 -> 2 -> 3 -> 4 -> 5
Output: 1 -> 5 -> 2 -> 4 -> 3
简单解决方案:
1) Initialize current node as head.
2) While next of current node is not null, do following
a) Find the last node, remove it from the end and insert it as next
of the current node.
b) Move current to next to next of current
上述简单解法的时间复杂度为 O(n 2 )，其中 n 为链表中的节点数。
更好的解决方案: 1)将给定链表的内容复制到一个向量。 2)通过交换两端的节点来重新排列给定的向量。 3)将修改后的向量复制回链表。 此方法的实施:https://ide.geeksforgeeks.org/1eGSEy 感谢阿鲁什·达美佳提出此方法。
高效解决方案:
1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.
该解决方案的时间复杂度为 0(n)。
下面是这个方法的实现。
java 描述语言
<script>
// Javascript program to rearrange
// linked list in place
// Linked List Class
// Head of the list
var head;
// Node Class
class Node
{
// Constructor to create
// a new node
constructor(d)
{
this.data = d;
this.next = null;
}
}
function printlist(node)
{
if (node == null)
{
return;
}
while (node != null)
{
document.write(node.data +
" -> ");
node = node.next;
}
}
function reverselist(node)
{
var prev = null,
curr = node, next;
while (curr != null)
{
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
node = prev;
return node;
}
function rearrange(node)
{
// 1) Find the middle point using
// tortoise and hare method
var slow = node, fast = slow.next;
while (fast != null &&
fast.next != null)
{
slow = slow.next;
fast = fast.next.next;
}
// 2) Split the linked list in
// two halves
// node1, head of first half
// 1 -> 2 -> 3
// node2, head of second half
// 4 -> 5
var node1 = node;
var node2 = slow.next;
slow.next = null;
// 3) Reverse the second half,
// i.e., 5 -> 4
node2 = reverselist(node2);
// 4) Merge alternate nodes
// Assign dummy Node
node = new Node(0);
// curr is the pointer to this
// dummy Node, which will be
// used to form the new list
var curr = node;
while (node1 != null ||
node2 != null)
{
// First add the element from
// first list
if (node1 != null)
{
curr.next = node1;
curr = curr.next;
node1 = node1.next;
}
// Then add the element from
// second list
if (node2 != null)
{
curr.next = node2;
curr = curr.next;
node2 = node2.next;
}
}
// Assign the head of the new
// list to head pointer
node = node.next;
}
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next =
new Node(4);
head.next.next.next.next =
new Node(5);
// Print original list
printlist(head);
// Rearrange list as per ques
rearrange(head);
document.write("<br/>");
// Print modified list
printlist(head);
// This code is contributed by gauravrajput1
</script>
输出:
1 -> 2 -> 3 -> 4 -> 5
1 -> 5 -> 2 -> 4 -> 3
时间复杂度:O(n) T3】辅助空间: O(1) 感谢高拉夫·阿希瓦尔提出上述方法。
另一种做法: 1。取两个指针 prev 和 curr，保存 head 和 head 的地址- >下一个。 2。比较他们的数据并交换。 之后，形成新的链表。
下面是实现:
java 描述语言
<script>
// Javascript code to rearrange linked list
// in place
class Node
{
constructor()
{
this.data;
this.next=null;
}
}
// Function for rearranging a linked list
// with high and low value.
function rearrange(head)
{
// Base case
if (head == null)
return null;
// Two pointer variable.
let prev = head, curr = head.next;
while (curr != null)
{
// Swap function for swapping data.
if (prev.data > curr.data)
{
let t = prev.data;
prev.data = curr.data;
curr.data = t;
}
// Swap function for swapping data
if (curr.next != null &&
curr.next.data > curr.data)
{
let t = curr.next.data;
curr.next.data = curr.data;
curr.data = t;
}
prev = curr.next;
if (curr.next == null)
break;
curr = curr.next.next;
}
return head;
}
// Function to display Node
// of linked list.
function display(head)
{
let curr = head;
while (curr != null)
{
document.write(curr.data+" ");
curr = curr.next;
}
}
// Function to insert a Node in
// the linked list at the beginning.
function push(head,k)
{
let tem = new Node();
tem.data = k;
tem.next = head;
head = tem;
return head;
}
// Driver code
let head = null;
head = push(head, 7);
head = push(head, 3);
head = push(head, 8);
head = push(head, 6);
head = push(head, 9);
head = rearrange(head);
display(head);
// This code is contributed by unknown2108
</script>
输出:
6 9 3 8 7
时间复杂度:O(n) T3】辅助空间: O(1) 感谢阿迪蒂亚提出这个方法。
另一种方法:(使用递归)
- 持有指向头节点的指针，并使用递归一直到最后一个节点
- 到达最后一个节点后，开始将最后一个节点交换到下一个头节点
- 将头指针移动到下一个节点
- 重复此操作，直到头部和最后一个节点相遇或相邻
- 一旦满足停止条件，我们需要丢弃左边的节点，以修复交换节点时在列表中创建的循环。
java 描述语言
<script>
// Javascript program to implement
// the above approach
// Creating the structure for node
class Node
{
// Function to create newNode
// in a linkedlist
constructor(val)
{
this.data = val;
this.next = null;
}
}
var left = null;
// Function to print the list
function printlist(head)
{
while (head != null)
{
document.write(head.data + " ");
if (head.next != null)
{
document.write("->");
}
head = head.next;
}
document.write("<br/>");
}
// Function to rearrange
function rearrange(head)
{
if (head != null)
{
left = head;
reorderListUtil(left);
}
}
function reorderListUtil(right)
{
if (right == null)
{
return;
}
reorderListUtil(right.next);
// We set left = null, when we reach
// stop condition, so no processing
// required after that
if (left == null)
{
return;
}
// Stop condition: odd case :
// left = right, even
// case : left.next = right
if (left != right &&
left.next != right)
{
var temp = left.next;
left.next = right;
right.next = temp;
left = temp;
}
else
{
// Stop condition , set null
// to left nodes
if (left.next == right)
{
// even case
left.next.next = null;
left = null;
}
else
{
// odd case
left.next = null;
left = null;
}
}
}
// Drivers Code
var head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next =
new Node(4);
head.next.next.next.next =
new Node(5);
// Print original list
printlist(head);
// Modify the list
rearrange(head);
// Print modified list
printlist(head);
// This code is contributed by aashish1995
</script>
输出:
1 ->2 ->3 ->4 ->5
1 ->5 ->2 ->4 ->3
请参考上的完整文章，就地重新排列给定的链表。了解更多详情！
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