Hashtable与双链表？

Question

Introduction to Algorithms (CLRS) states that a hash table using doubly linked lists is able to delete items more quickly than one with singly linked lists. Can anybody tell me what is the advantage of using doubly linked lists instead of single linked list for deletion in Hashtable implementation?

Answer 1

The confusion here is due to the notation in CLRS. To be consistent with the true question, I use the CLRS notation in this answer.

We use the hash table to store key-value pairs. The value portion is not mentioned in the CLRS pseudocode, while the key portion is defined as k .

In my copy of CLR (I am working off of the first edition here), the routines listed for hashes with chaining are insert, search, and delete (with more verbose names in the book). The insert and delete routines take argument x , which is the linked list element associated with key key[x] . The search routine takes argument k , which is the key portion of a key-value pair. I believe the confusion is that you have interpreted the delete routine as taking a key, rather than a linked list element.

Since x is a linked list element, having it alone is sufficient to do an O(1) deletion from the linked list in the h(key[x]) slot of the hash table, if it is a doubly-linked list . If, however, it is a singly-linked list, having x is not sufficient. In that case, you need to start at the head of the linked list in slot h(key[x]) of the table and traverse the list until you finally hit x to get its predecessor. Only when you have the predecessor of x can the deletion be done, which is why the book states the singly-linked case leads to the same running times for search and delete.

Additional Discussion

Although CLRS says that you can do the deletion in O(1) time, assuming a doubly-linked list, it also requires you have x when calling delete. The point is this: they defined the search routine to return an element x . That search is not constant time for an arbitrary key k . Once you get x from the search routine, you avoid incurring the cost of another search in the call to delete when using doubly-linked lists.

The pseudocode routines are lower level than you would use if presenting a hash table interface to a user. For instance, a delete routine that takes a key k as an argument is missing. If that delete is exposed to the user, you would probably just stick to singly-linked lists and have a special version of search to find the x associated with k and its predecessor element all at once.

Answer 2

I can think of one reason, but this isn't a very good one. Suppose we have a hash table of size 100. Now suppose values A and G are each added to the table. Maybe A hashes to slot 75. Now suppose G also hashes to 75, and our collision resolution policy is to jump forward by a constant step size of 80. So we try to jump to (75 + 80) % 100 = 55. Now, instead of starting at the front of the list and traversing forward 85, we could start at the current node and traverse backwards 20, which is faster. When we get to the node that G is at, we can mark it as a tombstone to delete it.

Still, I recommend using arrays when implementing hash tables.

Answer 3

Hashtable is often implemented as a vector of lists. Where index in vector is the key (hash).
If you don't have more than one value per key and you are not interested in any logic regarding those values a single linked list is enough. A more complex/specific design in selecting one of the values may require a double linked list.

Answer 4

Let's design the data structures for a caching proxy. We need a map from URLs to content; let's use a hash table. We also need a way to find pages to evict; let's use a FIFO queue to track the order in which URLs were last accessed, so that we can implement LRU eviction. In C, the data structure could look something like

struct node {
    struct node *queueprev, *queuenext;
    struct node **hashbucketprev, *hashbucketnext;
    const char *url;
    const void *content;
    size_t contentlength;
};
struct node *queuehead;  /* circular doubly-linked list */
struct node **hashbucket;

One subtlety: to avoid a special case and wasting space in the hash buckets, x->hashbucketprev points to the pointer that points to x . If x is first in the bucket, it points into hashbucket ; otherwise, it points into another node. We can remove x from its bucket with

x->hashbucketnext->hashbucketprev = x->hashbucketprev;
*(x->hashbucketprev) = x->hashbucketnext;

When evicting, we iterate over the least recently accessed nodes via the queuehead pointer. Without hashbucketprev , we would need to hash each node and find its predecessor with a linear search, since we did not reach it via hashbucketnext . (Whether that's really bad is debatable, given that the hash should be cheap and the chain should be short. I suspect that the comment you're asking about was basically a throwaway.)

Answer 5

If the items in your hashtable are stored in "intrusive" lists, they can be aware of the linked list they are a member of. Thus, if the intrusive list is also doubly-linked, items can be quickly removed from the table.

(Note, though, that the "intrusiveness" can be seen as a violation of abstraction principles...)

An example: in an object-oriented context, an intrusive list might require all items to be derived from a base class.

class BaseListItem {
  BaseListItem *prev, *next;

  ...

public: // list operations
  insertAfter(BaseListItem*);
  insertBefore(BaseListItem*);
  removeFromList();
};

The performance advantage is that any item can be quickly removed from its doubly-linked list without locating or traversing the rest of the list.

Answer 6

Unfortunately my copy of CLRS is in another country right now, so I can't use it as a reference. However, here's what I think it is saying:

Basically, a doubly linked list supports O(1) deletions because if you know the address of the item, you can just do something like:

x.left.right = x.right;
x.right.left = x.left;

to delete the object from the linked list, while as in a linked list, even if you have the address, you need to search through the linked list to find its predecessor to do:

pred.next = x.next

So, when you delete an item from the hash table, you look it up, which is O(1) due to the properties of hash tables, then delete it in O(1), since you now have the address.

If this was a singly linked list, you would need to find the predecessor of the object you wish to delete, which would take O(n).

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However:

I am also slightly confused about this assertion in the case of chained hash tables, because of how lookup works. In a chained hash table, if there is a collision, you already need to walk through the linked list of values in order to find the item you want, and thus would need to also find its predecessor.

But, the way the statement is phrased gives clarification: "If the hash table supports deletion, then its linked lists should be doubly linked so that we can delete an item quickly. If the lists were only singly linked, then to delete element x, we would first have to find x in the list T[h(x.key)] so that we could update the next attribute of x's predecessor."

This is saying that you already have element x, which means you can delete it in the above manner. If you were using a singly linked list, even if you had element x already, you would still have to find its predecessor in order to delete it.