4. Using Go language, explain how to allocate and release the storage space occupied by elements by linking all unoccupied slots into a free linked list inside the hash table. It is assumed that a slot can store a flag, an element plus one or two pointers. All dictionary and free linked list operations should have an expected running time of O(1). Does the free linked list need to be a doubly linked list? Or, is a singly linked list sufficient?
Wen Xin’s words:
In the Go language, you can use the built-in map type to implement a hash table, which uses hash tables and doubly linked lists internally to manage the storage and release of elements. Specifically, each bucket can store an element (key-value pair) and a pointer to the next element. When an element is inserted into the hash table, it is assigned to the corresponding slot and added to the end of the doubly linked list. When an element is deleted, its corresponding slot and linked list node are released.
Here’s how to allocate and free storage for elements in Go:
type Entry struct {<!-- -->
Key string
Value string
Next *Entry
}
type HashTable struct {<!-- -->
Buckets[]*Entry
}
//Initialize hash table
func NewHashTable(capacity int) *HashTable {<!-- -->
return & amp;HashTable{<!-- -->
Buckets: make([]*Entry, capacity),
}
}
//Insert element into hash table
func (h *HashTable) Insert(key, value string) {<!-- -->
hash := hash(key) % len(h.Buckets)
entry := & amp;Entry{<!-- -->Key: key, Value: value}
if h.Buckets[hash] == nil {<!-- -->
h.Buckets[hash] = entry
} else {<!-- -->
current := h.Buckets[hash]
for current.Next != nil {<!-- -->
current = current.Next
}
current.Next = entry
}
}
// Free up storage space by deleting elements
func (h *HashTable) Delete(key string) {<!-- -->
hash := hash(key) % len(h.Buckets)
if h.Buckets[hash] == nil {<!-- -->
return // key does not exist in the map
}
if h.Buckets[hash].Key == key {<!-- -->
h.Buckets[hash] = h.Buckets[hash].Next // remove the head of the list
} else {<!-- -->
current := h.Buckets[hash]
for current.Next != nil & amp; & amp; current.Next.Key != key {<!-- -->
current = current.Next
}
if current.Next == nil {<!-- -->
return // key does not exist in the list
}
current.Next = current.Next.Next // remove the node from the list
}
}
As for the question of whether the free linked list needs to be a doubly linked list, according to the map implementation of the Go language, it is necessary to use a doubly linked list. Because in a singly linked list, deleting a node requires traversing the entire linked list to find the node, while in a doubly linked list, deleting a node only requires finding the previous node. Therefore, for hash tables that undergo frequent insertion and deletion operations, using a doubly linked list can provide better performance.
Iflytek Spark, the code cannot run:
Within a hash table, the storage space occupied by elements can be allocated and released by linking all unoccupied slots into a free linked list. In order to achieve this function, we can use a doubly linked list to represent a free linked list. In this way, we can operate directly in the linked list when allocating and releasing elements, thus guaranteeing the expected running time of O(1).
The following is a simple Go language implementation:
package main
import (
"fmt"
)
type Node struct {<!-- -->
key int
value interface{<!-- -->}
prev *Node
next *Node
}
type HashTable struct {<!-- -->
size int
capacity int
buckets[]*Node
}
func NewHashTable(capacity int) *HashTable {<!-- -->
return & amp;HashTable{<!-- -->
size: 0,
capacity: capacity,
buckets: make([]*Node, capacity),
}
}
func (h *HashTable) hash(key int) int {<!-- -->
return key % h.capacity
}
func (h *HashTable) put(key int, value interface{<!-- -->}) {<!-- -->
index := h.hash(key)
node := h.buckets[index]
if node == nil {<!-- -->
node = & amp;Node{<!-- -->key: key, value: value}
h.buckets[index] = node
h.size++
} else {<!-- -->
for node != nil {<!-- -->
if node.key == key {<!-- -->
node.value = value
return
}
prev := node
node = node.next
}
newNode := & amp;Node{<!-- -->key: key, value: value, prev: prev}
prev.next = newNode
h.size++
}
}
func (h *HashTable) get(key int) (interface{<!-- -->}, bool) {<!-- -->
index := h.hash(key)
node := h.buckets[index]
for node != nil {<!-- -->
if node.key == key {<!-- -->
return node.value, true
}
node = node.next
}
return nil, false
}
func (h *HashTable) remove(key int) {<!-- -->
index := h.hash(key)
node := h.buckets[index]
for node != nil {<!-- -->
if node.key == key {<!-- -->
if node.prev != nil {<!-- -->
node.prev.next = node.next
} else {<!-- -->
h.buckets[index] = node.next
}
if node.next != nil {<!-- -->
node.next.prev = node.prev
}
h.size--
return
}
prev := node
node = node.next
}
}
func main() {<!-- -->
ht := NewHashTable(10)
ht.put(1, "one")
ht.put(2, "two")
ht.put(3, "three")
fmt.Println(ht.get(1)) // Output: one, true
fmt.Println(ht.get(4)) // Output: <nil>, false
ht.remove(2)
fmt.Println(ht.get(2)) // Output: <nil>, false
}
In this implementation, we use a doubly linked list to represent a free linked list. When we need to allocate a new element, we just insert a new node into the linked list. When we need to release an element, we only need to delete the corresponding node from the linked list. Since linked list operations have an expected running time of O(1), this implementation meets the requirements of the question.
360 Intelligent Brain:
Within the hash table, when all the slots in the hash table are occupied and no more new elements can be stored, we need to allocate and release the storage space occupied by the elements by linking all unoccupied slots into a free linked list.
The specific implementation is as follows:
1. First define a free linked list node structure, which contains a pointer to the previous node and a slot for storing elements. If the slot is already occupied by another node, the previous node will be empty.
type FreeNode struct {<!-- -->
Next *FreeNode // next node
}
2. Then define a FreeNode array to store all unoccupied slots, and use a hash function to calculate the slot where each element should be stored.
func newHashTable() *HashTable {<!-- -->
ht := & amp;HashTable{<!-- -->}
ht.buckets = make([]*Bucket, 0)
ht.hasher = make(hasher)
ht.freeNodes = make([]*FreeNode, 0)
// Calculate the size of the hash table and the size of the hash table
ht.bucketSize = 10
ht.hashSize = 3
ht.hasher.init(ht.bucketSize, ht.hashSize)
//Initialize hash table
ht.buckets = ht.hasher()
for i := 0; i < ht.buckets.Len(); i + + {<!-- -->
ht.buckets[i] = & amp;Bucket{<!-- -->}
}
//Initialize free linked list
ht.freeNodes = ht.freeNode()
for i := 0; i < ht.buckets.Len(); i + + {<!-- -->
ht.freeNode[i].Next = & amp;ht.freeNodes[ht.buckets.Index(i)]
}
ht.freeNodes = ht.freeNode[:ht.buckets.Len()]
return ht
}
3. The time complexity of operations such as inserting, deleting, and finding elements in a hash table is O(1).
4. If each slot only stores one element, then the free linked list needs to be a doubly linked list because it needs to be able to access elements in both directions. If each slot can store multiple elements, then a singly linked list is sufficient because only one-way access to the elements is required.
Wisdom spectrum clear words:
Within the hash table, elements can be allocated and released by linking all unoccupied slots into a free linked list. First, we need to create a one-way linked list structure to represent a free linked list. Since a one-way linked list only needs to maintain the tail pointer, the time complexity of inserting and deleting nodes is O(1). Assuming that a slot can store a flag, an element plus one or two pointers, we can design a simple singly linked list node structure:
type Bucket struct {<!-- -->
Mark bool
Key interface{<!-- -->}
Value interface{<!-- -->}
Next *Bucket
}
Among them, Mark is used to indicate whether the slot is occupied, Key and Value are used to store keys and values respectively, and Next is used to point to adjacent nodes.
Next, we implement the allocation and release operations of the hash table. When allocating, we start from the end of the free list and look for an unoccupied slot. If an unoccupied slot is found, it is removed from the free list and the new node it points to is set as the Next pointer. In addition, the Mark flag of the update slot is true, indicating that it is occupied. If an unoccupied slot cannot be found, new memory space needs to be allocated and a new node created.
When releasing, set the Mark flag of the slot to false, indicating that it is no longer occupied. Then reinsert it into the free linked list to make it a new node.
To achieve the desired running time of O(1), we need to maintain the tail pointer of the free linked list to quickly find and insert nodes during allocation and deallocation operations. Here’s a simple example:
func (b *Bucket) Alloc() (interface{<!-- -->}, bool) {<!-- -->
// Find unoccupied slots
for node := b.FreeList.Next; node != nil; node = node.Next {<!-- -->
if !node.Mark {<!-- -->
// Remove node and set marker
b.FreeList.Remove(node)
node.Mark = true
return node.Key, true
}
}
// No unoccupied slot found, allocate new memory
newNode := & amp;Bucket{<!-- -->
Mark: false,
Key: nil,
Value: nil,
Next: nil,
}
b.FreeList.InsertAtTail(newNode)
return newNode.Key, false
}
func (b *Bucket) Free(key interface{<!-- -->}) {<!-- -->
// Find the node with the specified key
node := b.Find(key)
if node == nil {<!-- -->
return
}
//Update node mark
node.Mark = false
}
func (b *Bucket) Find(key interface{<!-- -->}) *Bucket {<!-- -->
// Find the node with the specified key
node := b.FreeList.Next
for node != nil {<!-- -->
if node.Key == key {<!-- -->
return node
}
node = node.Next
}
return nil
}
//Auxiliary function for managing one-way linked lists
func (b *Bucket) Remove(node *Bucket) {<!-- -->
if node.Prev != nil {<!-- -->
node.Prev.Next = node.Next
} else {<!-- -->
b.FreeList.Head = node.Next
}
if node.Next != nil {<!-- -->
node.Next.Prev = node.Prev
}
}
func (b *Bucket) InsertAtTail(node *Bucket) {<!-- -->
b.FreeList.InsertAtTail(node)
}
b.FreeList in this example is a one-way linked list used to store unoccupied slots. In practical applications, the linked list structure can be adjusted according to needs to meet performance requirements in different scenarios. In short, a singly linked list is sufficient to meet the expected O(1) running time requirement in this scenario.
Tongyi Qianwen:
In the Go language, the method of allocating and releasing the storage space occupied by elements within a hash table is as follows:
First, assume that each slot has a flag to mark whether the slot is occupied. True if occupied; false otherwise. When inserting a new element, first find a free slot and set the slot to the occupied state. At this time, a free linked list pointing to all free slots needs to be maintained.
Secondly, when an element is released, the slot flag is set to unoccupied and the slot is added to the free linked list.
As for whether this free linked list should be one-way or two-way, it depends on the specific application scenario. Typically, doubly linked lists operate more efficiently because adjacent nodes need to be accessed when inserting and deleting elements. However, singly linked lists require less space and are more suitable for situations where memory is limited. Therefore, just choose according to the actual situation.
Additionally, to achieve the desired running time of O(1), a hash function can be used to map keys to slots in the hash table. In this way, the corresponding elements can be directly accessed, thereby improving the search speed. At the same time, in order to avoid hash conflicts, methods such as the chain address method can also be used to deal with collision problems.
Tiangong:
In a hash table, by linking all unoccupied slots into a free linked list, dynamic allocation and release of elements can be achieved, thereby improving the space utilization of the hash table. The specific implementation steps are as follows:
1. Define a structure, including a flag, an element and two pointers.
2. Define a hash table, containing an array and a pointer to the head of the free linked list.
3. When initializing the hash table, initialize all slots in the array to an empty structure, and point the head pointer of the linked list to the first slot of the array.
4. When an element needs to be inserted, first calculate the slot index of the element in the array. If the slot is already occupied, set the flag position in the structure of the slot to occupied, and point the head pointer of the linked list to the structure, thereby adding the structure to the linked list. Otherwise, set the flag of the slot to occupied and point the head pointer of the linked list to the slot.
5. When you need to delete an element, first calculate the slot index of the element in the array. If the slot is already occupied, set the flag of the slot to the idle state and point the head pointer of the linked list to the next slot. Otherwise, directly set the flag of the slot to the idle state.
6. When you need to obtain an element, first calculate the slot index of the element in the array. If the slot is already occupied, the element in the slot is returned directly. Otherwise, return empty.
For the implementation of a free linked list, either a doubly linked list or a singly linked list can be used. A doubly linked list can delete elements without modifying the pointers of the preceding and following elements, but it will increase storage space and maintenance costs. A singly linked list can save storage space and maintenance costs, but when deleting an element, you need to modify the pointers of the preceding and following elements. Therefore, the choice between singly linked list or doubly linked list depends on the specific implementation needs and performance requirements.
Ling Xiaozhi:
Inside the hash table, we can allocate and release elements by linking all unoccupied slots into a free linked list. It is assumed that each slot can store a flag, an element, and one or two pointers.
In order to have the expected O(1) running time for dictionary and free linked list operations, we can adopt the following steps:
1. Initialize the hash table: Create a hash table of initial size and mark all slots as unoccupied. At the same time, create a pointer to the head of the free linked list.
2. Allocate elements: When we need to allocate a new element, we take an unused slot from the free linked list. This is accomplished by marking the slot as occupied and adjusting the free list pointer.
3. Release an element: When we need to release an existing element, we mark the slot as unoccupied and add it to the head of the free linked list.
4. Find an element: To find an element, we can use a hash function to convert its key to an index and search in the corresponding slot. If the slot is occupied and the elements match, the element is returned. Otherwise, return Not Found.
As for whether a free linked list needs to be a doubly linked list or a singly linked list, it depends on the specific requirements for releasing elements and inserting elements. If you only need to perform insertion and release operations at the head of the linked list, then using a singly linked list is enough, which can maintain O(1) time complexity. If insertion and release operations need to be performed at any location in the linked list, a doubly linked list may be more suitable, but additional pointers are required to maintain the predecessor and successor nodes.
In short, implementing the allocation and release of elements through free linked lists inside a hash table is a common and efficient method that can meet the expected running time requirement of O(1).
chatgpt:
The number of times is limited.