用于合并排序链接列表的 Javascript 程序

合并排序通常是对链表进行排序的首选。链表的缓慢随机访问性能使得其他一些算法（如快速排序）表现不佳，而其他算法（如堆排序）则完全不可能。

设 head 为要排序的链表的第一个节点， headRef 为指向 head 的指针。请注意，我们需要在 MergeSort() 中引用 head，因为下面的实现会更改下一个链接以对链表（不是节点处的数据）进行排序，因此如果原始头部的数据不是链表中的最小值。

MergeSort(headRef)
1) If the head is NULL or there is only one element in the Linked List 
    then return.
2) Else divide the linked list into two halves.  
      FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
      MergeSort(a);
      MergeSort(b);
4) Merge the sorted a and b (using SortedMerge() discussed here) 
   and update the head pointer using headRef.
     *headRef = SortedMerge(a, b);

Javascript

 
// Javascript program to
// illustrate merge sorted
// of linkedList
 
 
    var head = null;
 
    // node a, b;
     class node {
            constructor(val) {
                this.val = val;
                this.next = null;
            }
        }
 
    function sortedMerge( a,  b)
    {
        var result = null;
        /* Base cases */
        if (a == null)
            return b;
        if (b == null)
            return a;
 
        /* Pick either a or b, and recur */
        if (a.val <= b.val) {
            result = a;
            result.next = sortedMerge(a.next, b);
        } else {
            result = b;
            result.next = sortedMerge(a, b.next);
        }
        return result;
    }
 
    function mergeSort( h) {
        // Base case : if head is null
        if (h == null || h.next == null) {
            return h;
        }
 
        // get the middle of the list
        var middle = getMiddle(h);
        var nextofmiddle = middle.next;
 
        // set the next of middle node to null
        middle.next = null;
 
        // Apply mergeSort on left list
        var left = mergeSort(h);
 
        // Apply mergeSort on right list
        var right = mergeSort(nextofmiddle);
 
        // Merge the left and right lists
        var sortedlist = sortedMerge(left, right);
        return sortedlist;
    }
 
    // Utility function to get the middle
    // of the linked list
    function getMiddle( head) {
        if (head == null)
            return head;
 
        var slow = head, fast = head;
 
        while (fast.next != null && fast.next.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    function push(new_data) {
        /* allocate node */
        var new_node = new node(new_data);
 
        /* link the old list off the new node */
        new_node.next = head;
 
        /* move the head to point to the new node */
        head = new_node;
    }
 
    // Utility function to print the linked list
    function printList( headref) {
        while (headref != null) {
            document.write(headref.val + " ");
            headref = headref.next;
        }
    }
 
     
  
        /*
         Let us create a unsorted linked
         list to test the functions
         created. The list shall be
         a: 2->3->20->5->10->15
         */
        push(15);
        push(10);
        push(5);
        push(20);
        push(3);
        push(2);
 
        // Apply merge Sort
        head = mergeSort(head);
        document.write("
 Sorted Linked List is:
");
        printList(head);
 
// This code contributed by umadevi9616
 

输出：

Sorted Linked List is: 
2 3 5 10 15 20

时间复杂度： O(n*log n)

空间复杂度： O(n*log n)

方法 2：这种方法更简单，使用 log n 空间。

合并排序（）：

1. 如果链表的大小为 1 则返回头部
2. 使用龟兔赛跑方法找到中间点
3. 将 mid 的下一个存储在 head2 中，即右子链表。
4. 现在使下一个中点为空。
5. 递归调用左右子链表的mergeSort()，存储左右链表的新头。
6. 给定左右子链表的新头参数，调用merge()，并存储合并后返回的最终头。
7. 返回合并链表的最终头。

合并（头1，头2）：

1. 取一个指针，说合并以将合并列表存储在其中并在其中存储一个虚拟节点。
2. 取一个指针 temp 并为其分配合并。
3. 如果 head1 的数据小于 head2 的数据，则将 head1 存储在 temp 的 next 中并将 head1 移动到 head1 的 next 中。
4. 否则将 head2 存储在 temp 的下一个并将 head2 移动到 head2 的下一个。
5. 将 temp 移到 temp 的下一个。
6. 重复步骤 3、4 和 5，直到 head1 不等于 null 并且 head2 不等于 null。
7. 现在将第一个或第二个链表的任何剩余节点添加到合并的链表中。
8. 返回合并后的下一个（将忽略虚拟并返回最终合并链表的头部）

Javascript

 
// JavaScript program for the above approach
 
// Node Class
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
    // Function to merge sort
    function mergeSort(head) {
        if (head.next == null)
            return head;
 
var mid = findMid(head);
var head2 = mid.next;
        mid.next = null;
var newHead1 = mergeSort(head);
var newHead2 = mergeSort(head2);
var finalHead = merge(newHead1, newHead2);
 
        return finalHead;
    }
 
    // Function to merge two linked lists
    function merge(head1,  head2) {
var merged = new Node(-1);
var temp = merged;
 
        // While head1 is not null and head2
        // is not null
        while (head1 != null && head2 != null) {
            if (head1.data < head2.data) {
                temp.next = head1;
                head1 = head1.next;
            } else {
                temp.next = head2;
                head2 = head2.next;
            }
            temp = temp.next;
        }
 
        // While head1 is not null
        while (head1 != null) {
            temp.next = head1;
            head1 = head1.next;
            temp = temp.next;
        }
 
        // While head2 is not null
        while (head2 != null) {
            temp.next = head2;
            head2 = head2.next;
            temp = temp.next;
        }
        return merged.next;
    }
 
    // Find mid using The Tortoise and The Hare approach
    function findMid(head) {
var slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    // Function to print list
    function printList(head) {
        while (head != null) {
            document.write(head.data + " ");
            head = head.next;
        }
    }
 
    // Driver Code
     
var head = new Node(7);
var temp = head;
        temp.next = new Node(10);
        temp = temp.next;
        temp.next = new Node(5);
        temp = temp.next;
        temp.next = new Node(20);
        temp = temp.next;
        temp.next = new Node(3);
        temp = temp.next;
        temp.next = new Node(2);
        temp = temp.next;
 
        // Apply merge Sort
        head = mergeSort(head);
        document.write("Sorted Linked List is: <br/>");
        printList(head);
 
// This code contributed by gauravrajput1
 

输出：

Sorted Linked List is: 
2 3 5 7 10 20 

时间复杂度：O(n*log n)

空间复杂度： O(log n)

有关详细信息，请参阅有关链接列表的合并排序的完整文章！