用于从线段链接列表中删除中间点的 Javascript 程序

给定一个坐标链表，其中相邻点形成垂直线或水平线。从链表中删除位于水平或垂直线中间的点。
例子：

Input:  (0,10)->(1,10)->(5,10)->(7,10)
                                  |
                                (7,5)->(20,5)->(40,5)
Output: Linked List should be changed to following
        (0,10)->(7,10)
                  |
                (7,5)->(40,5) 
The given linked list represents a horizontal line from (0,10) 
to (7, 10) followed by a vertical line from (7, 10) to (7, 5), 
followed by a horizontal line from (7, 5) to (40, 5).

Input:     (2,3)->(4,3)->(6,3)->(10,3)->(12,3)
Output: Linked List should be changed to following
    (2,3)->(12,3) 
There is only one vertical line, so all middle points are removed.

来源：微软面试体验

这个想法是跟踪当前节点、下一个节点和下一个下一个节点。当下一个节点与下一个节点相同时，继续删除下一个节点。在这个完整的过程中，我们需要密切关注指针的移动并检查 NULL 值。
以下是上述想法的实现。

Javascript

 
// Javascript program to remove middle
// points in a linked list of
// line segments,
var  head; // head of list
 
    /* Linked list Node */
    class Node {
 
constructor(x , y) {
            this.x = x;
            this.y = y;
            this.next = null;
        }
    }
 
    // This function deletes middle
    // nodes in a sequence of
    // horizontal and vertical line
    // segments represented
    // by linked list.
    function deleteMiddle() {
        // If only one node or no
        // node...Return back
        if (head == null || head.next == null ||
        head.next.next == null)
            return head;
 
var Next = head.next;
var NextNext = Next.next;
 
        // check if this is vertical or
        // horizontal line
        if (head.x == Next.x) {
            // Find middle nodes with same
            // value as x and
            // delete them.
            while (NextNext != null &&
            Next.x == NextNext.x)
            {
                head.next = Next.next;
                Next.next = null;
 
                // Update NextNext for the next iteration
                Next = NextNext;
                NextNext = NextNext.next;
            }
        }
 
        // if horizontal
        else if (head.y == Next.y) {
            // find middle nodes with same value as y and
            // delete them
            while (NextNext != null &&
            Next.y == NextNext.y) {
                head.next = Next.next;
                Next.next = null;
 
                // Update NextNext for the next iteration
                Next = NextNext;
                NextNext = NextNext.next;
            }
        }
 
        // Adjacent points should have same x or same y
        else {
            document.write("Given list is not valid");
            return null;
        }
 
        // recur for other segment
 
        // temporarily store the head and move head forward.
var temp = head;
        head = head.next;
 
        // call deleteMiddle() for next segment
        this.deleteMiddle();
 
        // restore head
        head = temp;
 
        // return the head
        return head;
    }
 
    /*
     Given a reference (pointer to pointer) to
     the head of a list and an int, push
     a new node on the front of the list.
     */
    function push(x , y) {
        /*
         1 & 2: Allocate the Node & Put in the data
         */
var new_node = new Node(x, y);
 
        /* 3. Make next of new Node as head */
        new_node.next = head;
 
        /* 4. Move the head to point to new Node */
        head = new_node;
    }
 
    function printList() {
var temp = head;
        while (temp != null) {
            document.write("(" + temp.x + "," +
            temp.y + ")->");
            temp = temp.next;
        }
        document.write("<br/>");
    }
 
    /* Driver program to test above functions */
     
 
        push(40, 5);
        push(20, 5);
        push(10, 5);
        push(10, 8);
        push(10, 10);
        push(3, 10);
        push(1, 10);
        push(0, 10);
 
        document.write("Given list<br/>");
        printList();
 
        if (deleteMiddle() != null) {
            document.write("Modified Linked List is<br/>");
            printList();
        }
 
// This code contributed by gauravrajput1
 

输出：

Given Linked List:
(0,10)-> (1,10)-> (3,10)-> (10,10)-> (10,8)-> (10,5)-> (20,5)-> (40,5)->
Modified Linked List:
(0,10)-> (10,10)-> (10,5)-> (40,5)-> 

上述解决方案的时间复杂度为 O(n)，其中 n 是给定链表中的节点数。
锻炼：
上面的代码是递归的，针对同样的问题写一个迭代代码。请参阅下面的解决方案。
删除线段链表中点的迭代方法
请参阅完整的文章给定线段的链表，删除中间点以获取更多详细信息！