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  1. Prepare
  2. Data Structures
  3. Linked Lists
  4. Merge two sorted linked lists
  5. Discussions

Merge two sorted linked lists

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977 Discussions

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  • cute_n_charming
     + 0 comments

    function mergeLists(head1, head2) { let newNode = new SinglyLinkedListNode(); let tail = newNode;

    while(head1 !==null && head2 !== null){
    if(head1.data < head2.data){
        tail.next = head1;
        head1= head1.next;
    }else if(head2.data <= head1.data){
        tail.next = head2;
        head2 = head2.next;
    }
    tail = tail.next;


    if(head1 !==null){
        tail.next = head1;
    }else if(head2 !== null){
        tail.next = head2;
    }
}
    return newNode.next;

    }

  • namnika_janband1
     + 0 comments

    javascript:

    function mergeLists(head1, head2) {
    let newnode = new SinglyLinkedListNode(0, null)
    let tempnode = newnode
    
    while(head1 !== null && head2 !== null ){
        if(head1.data < head2.data){
            tempnode.next = head1
            head1 = head1.next //inc of head1 linked list
        }else{
            tempnode.next = head2
            head2 = head2.next  //inc of head2 linked list
        }
        tempnode = tempnode.next //inc of newnode
    }
    
    tempnode.next = head1 || head2 // if either head1 is null then tempnode will update with head2 or vice-versa
    
    return newnode.next
}

  • burham
     + 0 comments

    Javascript solution:

    function mergeLists(head1, head2) {
  let head, list = getItems(head2).split(',')
  .concat(getItems(head1).split(','))
  .sort((a,b) => b-a)
	
  for(let i =0; i<list.length; i++ ){
    let res = new SinglyLinkedListNode(list[i])
    res.next = head
    head = res
  }
   return head
}

function getItems(item){
  let list = []
  Object.entries(item).forEach((a,b) =>{
    if(typeof(a[1]) == 'object' && a[1] != null) list.push(getItems(a[1]))
    if(typeof(a[1]) == 'number') list.push(a[1])
  })
  return list.toString()
}

  • pvolpe
     + 0 comments

    Two ways

    def mergeLists(head1, head2):
    l, result = [], SinglyLinkedList()
    while head1:
        l.append(head1.data)
        head1 = head1.next
    while head2:
        l.append(head2.data)
        head2 = head2.next
    for data in sorted(l):
        result.insert_node(data)
    return result.head

def mergeLists(head1, head2):
    result = SinglyLinkedList()
    while head1 or head2:
        if not head1:
            result.insert_node(head2.data)
            head2 = head2.next
        elif not head2:
            result.insert_node(head1.data)
            head1 = head1.next
        elif head1 and head2:
            if head1.data <= head2.data:
                result.insert_node(head1.data)
                head1 = head1.next
            else:
                result.insert_node(head2.data)
                head2 = head2.next
    return result.head

  • abraham_balisbis
     + 0 comments

    Solution in C++:

    SinglyLinkedListNode* mergeLists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
    
    SinglyLinkedListNode* head = NULL;
    if(head1 == NULL)
        return head2;
    else if(head2 == NULL)
        return head1;
    else
    {
        if(head1->data <= head2->data)
        {
            head = head1;
            head1 = head1->next;
        }
        else
        {
            head = head2;
            head2 = head2->next;
        }
        
        SinglyLinkedListNode* ptr;
        ptr = head;
        while(head1 != NULL || head2 != NULL)
        {
            if(head1 != NULL && head2 != NULL)
            {
                if(head1->data <= head2->data)
                {
                    ptr->next = head1;
                    head1 = head1->next;
                }
                else
                {
                    ptr->next = head2;
                    head2 = head2->next;
                }
            }
            else if(head1 != NULL)
            {
                ptr->next = head1;
                head1 = head1->next;
            }
            else if(head2 != NULL)
            {
                ptr->next = head2;
                head2 = head2->next;
            }
            ptr = ptr->next;
        }
    }        
    return head;
}