Delete the repeating elements in the sorting list (simple questions)

Before doing this topic , You need to have some prerequisite knowledge about the linked list first , Make one here JavaScript There is a supplement to the linked list .

The first thing to be clear is ：

Although the linked list looks like an array , But linked lists and arrays are different ！！！ Not a data structure ！

The positions of linked list elements in memory are not continuous （ Arrays are contiguous storage ）, Each element consists of a node that stores the element itself and a reference to the next element ( Also called pointer or link ) form .

For a more convenient understanding , Here is an example diagram ：

（ This is a 3 Single linked list of elements ）

Because of this characteristic , The linked list is being deleted , It is much more convenient to insert than array ！item Save the data ,next Save the reference of the next linked list .

Inserting a linked list is easier than an array , Here is an example diagram ：

What we need to do is to specify the location of the linked list element to be inserted into the linked list next The value refers to the element that needs to be inserted data value , While inserting elements next Value points to the next element of the inserted element data value ;

Code shows ：

function insert ( newElement , item ) {
    var newNode = new Node( newElement );
    var currNode = this.find( item );
    newNode.next = currNode.next;
    currNode.next = newNode;
}

  Deleting a linked list is very similar to inserting a linked list , So just give an example diagram ：

The following is a description of the title ：
Given the header of a sorted linked list  head ,  Delete all duplicate elements , Make each element appear only once  . return   Sorted linked list  .

Example diagram description ：

 

  Their thinking ：

Find repeating elements

var deleteDuplicates = function(head) {
    var fir=head
    while(fir && fir.next){
        if(fir.val ==fir.next.val){
            fir.next=fir.next.next
        }else{
            fir=fir.next
        }
    }
    return head
};