在此之前,我对我的经验不足深表歉意,这些都是难以理解的高级概念。据我了解,线性探测是循环的,直到找到一个空单元格它才会停止。
但是我不确定如何实现它。一些关于如何的示例将不胜感激。再次抱歉,我没有足够的经验,我不是经过审查的程序员,我的学习过程非常缓慢。
public boolean ContainsElement(V element)
{
for(int i = 0; i < capacity; i++)
{
if(table[i] != null)
{
LinkedList<Entry<K, V>> bucketMethod = table[i];
for(Entry<K, V> entry : bucketMethod)
{
if(entry.getElement().equals(element))
{
return true;
}
}
}
}
return false;
}
答案 0 :(得分:1)
这是一个基于Wikipedia article for open addressing中的伪代码示例的工作哈希表。
我认为Wikipedia示例与我的示例之间的主要区别是:
hashCode()
)进行模数运算的方式,对%
进行了一些处理。remove
方法中的逻辑,因为Java没有goto
。否则,它或多或少只是直接翻译。
package mcve;
import java.util.*;
import java.util.stream.*;
public class OAHashTable {
private Entry[] table = new Entry[16]; // Must be >= 4. See findSlot.
private int size = 0;
public int size() {
return size;
}
private int hash(Object key) {
int hashCode = Objects.hashCode(key)
& 0x7F_FF_FF_FF; // <- This is like abs, but it works
// for Integer.MIN_VALUE. We do this
// so that hash(key) % table.length
// is never negative.
return hashCode;
}
private int findSlot(Object key) {
int i = hash(key) % table.length;
// Search until we either find the key, or find an empty slot.
//
// Note: this becomes an infinite loop if the key is not already
// in the table AND every element in the array is occupied.
// With the resizing logic (below), this will only happen
// if the table is smaller than length=4.
while ((table[i] != null) && !Objects.equals(table[i].key, key)) {
i = (i + 1) % table.length;
}
return i;
}
public Object get(Object key) {
int i = findSlot(key);
if (table[i] != null) { // Key is in table.
return table[i].value;
} else { // Key is not in table
return null;
}
}
private boolean tableIsThreeQuartersFull() {
return ((double) size / (double) table.length) >= 0.75;
}
private void resizeTableToTwiceAsLarge() {
Entry[] old = table;
table = new Entry[2 * old.length];
size = 0;
for (Entry e : old) {
if (e != null) {
put(e.key, e.value);
}
}
}
public void put(Object key, Object value) {
int i = findSlot(key);
if (table[i] != null) { // We found our key.
table[i].value = value;
return;
}
if (tableIsThreeQuartersFull()) {
resizeTableToTwiceAsLarge();
i = findSlot(key);
}
table[i] = new Entry(key, value);
++size;
}
public void remove(Object key) {
int i = findSlot(key);
if (table[i] == null) {
return; // Key is not in the table.
}
int j = i;
table[i] = null;
--size;
while (true) {
j = (j + 1) % table.length;
if (table[j] == null) {
break;
}
int k = hash(table[j].key) % table.length;
// Determine if k lies cyclically in (i,j]
// | i.k.j |
// |....j i.k.| or |.k..j i...|
if ( (i<=j) ? ((i<k)&&(k<=j)) : ((i<k)||(k<=j)) ) {
continue;
}
table[i] = table[j];
i = j;
table[i] = null;
}
}
public Stream<Entry> entries() {
return Arrays.stream(table).filter(Objects::nonNull);
}
@Override
public String toString() {
return entries().map(e -> e.key + "=" + e.value)
.collect(Collectors.joining(", ", "{", "}"));
}
public static class Entry {
private Object key;
private Object value;
private Entry(Object key, Object value) {
this.key = key;
this.value = value;
}
public Object getKey() { return key; }
public Object getValue() { return value; }
}
public static void main(String[] args) {
OAHashTable t = new OAHashTable();
t.put("A", 1);
t.put("B", 2);
t.put("C", 3);
System.out.println("size = " + t.size());
System.out.println(t);
t.put("X", 4);
t.put("Y", 5);
t.put("Z", 6);
t.remove("C");
t.remove("B");
t.remove("A");
t.entries().map(e -> e.key)
.map(key -> key + ": " + t.get(key))
.forEach(System.out::println);
}
}
答案 1 :(得分:-1)
java.util.Map的java.util.HashMap实现内部提供了线性探测,即HashMap可以解决哈希表中的冲突。