Wrote the below function to merge two linkedlists:
var mergeTwoLists = function(l1, l2) {
current = l1.head;
if (current === null){
l1.head = l2.head; //Have to use l1.head instead of current since if we assign current again to l2.head the current will just start pointing to l2.head and lose reference to l1.
}
else{
while( current.next != null){
current = current.next;
}
current.next = l2.head;
}
return l1;
};
Creating two linked lists:
let l1 = new linkedlist();
let l2 = new linkedlist();
l1.insert(1);
l1.insert(2);
l1.insert(4);
l2.insertEnd(1);
l2.insertEnd(3);
l2.insertEnd(4);
Now i called the function twice:
let l3 = mergeTwoLists(l1,l2);
console.log(l3.show());
let l4 = mergeTwoLists(l1,l2);
console.log(l4.show())
The first show outputs the expected 4 2 1 1 3 4. However, the second call goes on an infinite loop and keeps outputting 4 2 1 1 3 4 1 3 4 1 3 4 1 3 4 ........
Why is this happening?
mergeTwoLists doesn't make copies of the list nodes. So after the first merge, the last node of l1 is the same as the last node of l2. When you merge them again, the next property of the last node in l2 now points to l2.head, resulting in a circular list, just as if you'd done mergeTwoLists(l2, l2).
If you don't want this to happen, you should define a copyList() function, and use
current.next = copyList(l2).head;