我正在用Java编写自己的自定义HashMap实现。以下是我的意见。
public class Entry<K,V> {
private final K key;
private V value;
private Entry<K,V> next;
public Entry(K key, V value, Entry<K,V> next) {
this.key = key;
this.value = value;
this.next = next;
}
public V getValue() {
return value;
}
public void setValue(V value) {
this.value = value;
}
public Entry<K, V> getNext() {
return next;
}
public void setNext(Entry<K, V> next) {
this.next = next;
}
public K getKey() {
return key;
}
}
public class MyCustomHashMap<K,V> {
private int DEFAULT_BUCKET_COUNT = 10;
private Entry<K,V>[] buckets;
public MyCustomHashMap() {
buckets = new Entry[DEFAULT_BUCKET_COUNT];
for (int i = 0;i<DEFAULT_BUCKET_COUNT;i++)
buckets[i] = null;
}
public void put(K key,V value){
/**
* This is the new node.
*/
Entry<K,V> newEntry = new Entry<K,V>(key, value, null);
/**
* If key is null, then null keys always map to hash 0, thus index 0
*/
if(key == null){
buckets[0] = newEntry;
}
/**
* get the hashCode of the key.
*/
int hash = hash(key);
/**
* if the index does of the bucket does not contain any element then assign the node to the index.
*/
if(buckets[hash] == null) {
buckets[hash] = newEntry;
} else {
/**
* we need to traverse the list and compare the key with each of the keys till the keys match OR if the keys does not match then we need
* to add the node at the end of the linked list.
*/
Entry<K,V> previous = null;
Entry<K,V> current = buckets[hash];
while(current != null) {
boolean done = false;
while(!done) {
if(current.getKey().equals(key)) {
current.setValue(value);
done = true; // if the keys are same then replace the old value with the new value;
} else if (current.getNext() == null) {
current.setNext(newEntry);
done = true;
}
current = current.getNext();
previous = current;
}
}
previous.setNext(newEntry);
}
}
public V getKey(K key) {
int hash = hash(key);
if(buckets[hash] == null) {
return null;
} else {
Entry<K,V> temp = buckets[hash];
while(temp != null) {
if(temp.getKey().equals(key))
return temp.getValue(); // returns value corresponding to key.
temp = temp.getNext();
}
return null; //return null if key is not found.
}
}
public void display() {
for(int i = 0; i < DEFAULT_BUCKET_COUNT; i++) {
if(buckets[i] != null) {
Entry<K,V> entry = buckets[i];
while(entry != null){
System.out.print("{"+entry.getKey()+"="+entry.getValue()+"}" +" ");
entry=entry.getNext();
}
}
}
}
public int bucketIndexForKey(K key) {
int bucketIndex = key.hashCode() % buckets.length;
return bucketIndex;
}
/**
*
* @param key
* @return
*/
private int hash(K key){
return Math.abs(key.hashCode()) % buckets.length;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
MyCustomHashMap<String, Integer> myCustomHashMap = new MyCustomHashMap<String, Integer>();
myCustomHashMap.put("S", 22);
myCustomHashMap.put("S", 1979);
myCustomHashMap.put("V", 5);
myCustomHashMap.put("R", 31);
System.out.println("Value corresponding to key R: "+myCustomHashMap.getKey("R"));
System.out.println("Value corresponding to key V: "+myCustomHashMap.getKey("V"));
System.out.println("Displaying the contents of the HashMap:: ");
myCustomHashMap.display();
}
}
1）我觉得put（K键，V值）有些瑕疵。请做好验证，让我知道这里有什么问题。在输入相同的密钥时，它给我错误的结果。我还没有测试它有不同键的碰撞情况。
2）据说我们重新哈希hashCode，以便消除hashCode的错误实现。我该怎么做呢，因为如果我给key的hashCode，即hash（key.hashCode（））那么它就不能算，因为它不能计算int的hashCode。怎么做？
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由于 SID
答案 0 :(得分：3)
您错误地处理空键：
if(key == null){
buckets[0] = newEntry;
}
buckets[0]可能已包含条目，在这种情况下，您将丢失这些条目。
以下循环存在一些问题：
Entry<K,V> previous = null;
Entry<K,V> current = buckets[hash];
while(current != null) {
boolean done = false;
while(!done) {
if(current.getKey().equals(key)) {
current.setValue(value);
done = true;
} else if (current.getNext() == null) {
current.setNext(newEntry);
done = true;
}
current = current.getNext();
previous = current; // you are not really setting previous to
// to the previous Entry in the list - you
// are setting it to the current Entry
}
}
previous.setNext(newEntry); // you don't need this statement. You
// already have a statement inside the
// loop that adds the new Entry to the list
看起来删除与previous相关的任何语句都会修复此循环。
编辑：
正如kolakao评论的那样，为了使您的实施更有效（即get和put需要预期的固定时间），您必须在条目数超过时调整HashMap的大小一些阈值（为了使每个桶中的平均条目数由常量绑定）。
据说我们重新散列hashCode，以便消除hashCode的错误实现。我该怎么做，因为如果我给key的hashCode，即hash（key.hashCode（）），那么它不能，因为它无法计算int的hashCode。怎么做？
重新散列的想法不涉及为密钥的hashCode调用hashCode。它涉及对key.hashCode()获得的值运行一些硬编码函数。
例如，在Java 7的{{1}}实现中，使用了以下函数：
HashMap然后你用它：
static int hash(int h) {
// This function ensures that hashCodes that differ only by
// constant multiples at each bit position have a bounded
// number of collisions (approximately 8 at default load factor).
h ^= (h >>> 20) ^ (h >>> 12);
return h ^ (h >>> 7) ^ (h >>> 4);
}