You can also take a look at llist python package, which provides some useful features that deque does not. There are not only doubly linked lists, but also single linked lists data structure in that package. IMHO, one of the biggest advantages of this package is the ability to store a reference to the llist … Read more
That question gets asked a lot. When I was asked it in my interviews many years ago, I reasoned as follows: a singly-linked list is essentially a stack. Reversing a linked list is therefore a trivial operation on stacks: newList = emptyList; while(!oldList.IsEmpty()) newList.Push(oldList.Pop()); Now all you have to do is implement IsEmpty and Push … Read more
Node next = tmp.next; while(tmp != null){ So what happens when tmp == null? You almost got it, though. Node before = null; Node tmp = head; while (tmp != null) { Node next = tmp.next; tmp.next = before; before = tmp; tmp = next; } head = before; Or in nicer (?) naming: Node … Read more
Step1: Proceed in the usual way, you will use to find the loop, i.e. Have two pointers, increment one in single step and other in two steps, If they both meet in sometime, there is a loop. Step2: Freeze one pointer where it was and increment the other pointer in one step counting the steps … Read more
Yes* Yes, usually (and in your case) it does break out of the loop and returns from the method. An Exception One exception is that if there is a finally block inside the loop and surrounding the return statement then the code in the finally block will be executed before the method returns. The finally … Read more
Node MergeLists(Node list1, Node list2) { if (list1 == null) return list2; if (list2 == null) return list1; if (list1.data < list2.data) { list1.next = MergeLists(list1.next, list2); return list1; } else { list2.next = MergeLists(list2.next, list1); return list2; } }
Any alternative? No, this is as simple as it gets, and there’s no fundamentally-different way of doing it. This algorithm is already O(n) time, and you can’t get any faster than that, as you must modify every node. It looks like your code is on the right track, but it’s not quite working in the … Read more