在LinkedHashMap中以O(1)检查两个不同键的顺序。

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英文:

Check the ordering of 2 different keys in LinkedHashMap in O(1)

问题

在LinkedHashMap中,键的顺序是根据它们插入到映射中的时间来排序的。

例如,考虑以下映射。

LinkedHashMap<Integer, Integer> map = new LinkedHashMap<>();
map.put(1, 1);
map.put(2, 3);
map.put(7, 4);
map.put(5, 6);

所以,我对LinkedHashMap的理解是,键以双向链表的形式存储。

因此,在这个示例中,键的顺序将是1<->2<->7<->5

现在,我的需求是这样的。给定来自这个映射的两个键,是否可以检查哪一个先插入,哪一个后插入?

例如,一个像下面这样的函数:

public boolean isFirst(int key1, int key2) {
    // 用于计算key1是否在key2之前的逻辑
}

现在,可以轻松地通过维护另一个映射以获取顺序并进行比较来在O(n)中完成上述函数。

我想在O(1)中执行此操作。如果可能的话,我应该维护一些其他数据结构来处理映射中键的插入、删除和更新吗?

英文:

In a LinkedHashMap, the keys are ordered by when they were inserted into the map.

For example, consider the following map.

LinkedHashMap&lt;Integer,Integer&gt; map = new LinkedHashMap&lt;&gt;();
map.put(1, 1);
map.put(2, 3);
map.put(7, 4);
map.put(5, 6);

So, my understanding of LinkedHashMap is that the keys are stored in the form a Doubly Linked List.

So, in this example, the keys would be ordered as 1&lt;-&gt;2&lt;-&gt;7&lt;-&gt;5

Now, my requirement is this. Given 2 keys from this map, is it possible to check which one came first or last?

For example, a function like the following

public boolean isFirst(int key1,int key2){
// logic to compute if key1 is first
}

Now, the above function can be done in O(n) easily by maintaining another map to get the order and compare it.

I want to do this operation in O(1). If it is possible, then should I maintain some other data structure to handle insertion, removal and updation of keys in the map ?

答案1

得分: 2

你可以创建一个自定义类,它扩展了 LinkedHashMap,并添加了自定义方法 isFirst。在内部,我们保留了一个单独的映射,就像你已经分析过的那样,其中有一个 timestamp 变量,它充当了时间戳,表示键添加的秒数。

如果第一个键的时间戳小于第二个键的时间戳,我们返回 true,否则返回 false

类片段:

class CustomMap<K,V> extends LinkedHashMap<K,V>{
    private Map<K,Integer> ord_map;
    private int timestamp = 0;
    CustomMap(){
        super();
        ord_map = new HashMap<>();
    }
    
    public V put(K key,V val){
        ord_map.put(key,timestamp++);
        return super.put(key,val);
    }
    
    public V remove(Object key){
        ord_map.remove(key);
        return super.remove(key);
    }
    
    public boolean isFirst(K key1,K key2) throws Exception{
        if(!ord_map.containsKey(key1)) throw new Exception(key1 + " 不存在。");
        if(!ord_map.containsKey(key2)) throw new Exception(key2 + " 不存在。");
        
        return ord_map.get(key1) < ord_map.get(key2);
    }
}

驱动代码:

public class Solution{
    public static void main(String[] args) {
        CustomMap<Integer,Integer> cmap = new CustomMap<>();
        try{
            cmap.put(1, 1);
            cmap.put(5, 6);
            cmap.put(2, 3);
            cmap.put(7, 4);
            System.out.println(cmap.isFirst(5,2));    
            cmap.remove(5);
            cmap.put(5,99);
            System.out.println(cmap.isFirst(5,2));  
        }catch(Exception e){
            System.out.println(e.getMessage());
        }
    }
}
英文:

You can create a custom class that extends LinkedHashMap and add the custom method isFirst. Internally, we keep a separate map as you already analyzed with a timestamp variable which acts like a timestamp for us(the second at which the key was added).

If the first key's timestamp is less than the second ones, we return true , else we return false.

Class Snippet:

class CustomMap&lt;K,V&gt; extends LinkedHashMap&lt;K,V&gt;{
    private Map&lt;K,Integer&gt; ord_map;
    private int timestamp = 0;
    CustomMap(){
        super();
        ord_map = new HashMap&lt;&gt;();
    }
    
    public V put(K key,V val){
        ord_map.put(key,timestamp++);
        return super.put(key,val);
    }
    
    public V remove(Object key){
        ord_map.remove(key);
        return super.remove(key);
    }
    
    public boolean isFirst(K key1,K key2) throws Exception{
        if(!ord_map.containsKey(key1)) throw new Exception(key1 + &quot; does not exist.&quot;);
        if(!ord_map.containsKey(key2)) throw new Exception(key2 + &quot; does not exist.&quot;);
        
        return ord_map.get(key1) &lt; ord_map.get(key2);
    }
}

Driver Code:

public class Solution{
    public static void main(String[] args) {
        CustomMap&lt;Integer,Integer&gt; cmap = new CustomMap&lt;&gt;();
        try{
            cmap.put(1, 1);
            cmap.put(5, 6);
            cmap.put(2, 3);
            cmap.put(7, 4);
            System.out.println(cmap.isFirst(5,2));    
            cmap.remove(5);
            cmap.put(5,99);
            System.out.println(cmap.isFirst(5,2));  
        }catch(Exception e){
            System.out.println(e.getMessage());
        }
        
    }
}

答案2

得分: 1

以下是翻译好的部分:

有一种基于HashMap的索引解决方案,但在填充索引映射时,它的时间和内存复杂度都是O(N)。

import java.util.*;

public class Solution {
    static int index = 0;
    
    static Map<Integer, Integer> indexMap = new HashMap<>();
    
    static void initIndexes(LinkedHashMap<Integer, Integer> map) {
        indexMap.clear();
        index = 0;
        map.keySet().forEach(x -> {indexMap.put(x, index++);});
    }
    
    static boolean isKeyBefore(int key1, int key2) {
        return indexMap.getOrDefault(key1, Integer.MAX_VALUE) < 
        indexMap.getOrDefault(key2, -1);
    }

    public static void main(String []args) {
        LinkedHashMap<Integer,Integer> map = new LinkedHashMap<>();
        map.put(1, 1);
        map.put(2, 3);
        map.put(7, 4);
        map.put(5, 6);
        
        initIndexes(map);
            
        System.out.println(indexMap);
        
        System.out.println(isKeyBefore(2, 5)); // true
        System.out.println(isKeyBefore(7, 1)); // false
        System.out.println(isKeyBefore(5, 3)); // key2 is not in the indexMap
        System.out.println(isKeyBefore(4, 2)); // key1 is not in the indexMap
        System.out.println(isKeyBefore(6, 8)); // both keys are not in the indexMap
        
    }
}
英文:

There could be a solution based on HashMap for indexes, but it has O(N) complexity both in time and memory to populate the index map.

import java.util.*;
public class Solution {
static int index = 0;
static Map&lt;Integer, Integer&gt; indexMap = new HashMap&lt;&gt;();
static void initIndexes(LinkedHashMap&lt;Integer, Integer&gt; map) {
indexMap.clear();
index = 0;
map.keySet().forEach(x -&gt; {indexMap.put(x, index++);});
}
static boolean isKeyBefore(int key1, int key2) {
return indexMap.getOrDefault(key1, Integer.MAX_VALUE) &lt; 
indexMap.getOrDefault(key2, -1);
}
public static void main(String []args) {
LinkedHashMap&lt;Integer,Integer&gt; map = new LinkedHashMap&lt;&gt;();
map.put(1, 1);
map.put(2, 3);
map.put(7, 4);
map.put(5, 6);
initIndexes(map);
System.out.println(indexMap);
System.out.println(isKeyBefore(2, 5)); // true
System.out.println(isKeyBefore(7, 1)); // false
System.out.println(isKeyBefore(5, 3)); // key2 is not in the indexMap
System.out.println(isKeyBefore(4, 2)); // key1 is not in the indexMap
System.out.println(isKeyBefore(6, 8)); // both keys are not in the indexMap
}
}

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  • 本文由 发表于 2020年7月23日 02:11:20
  • 转载请务必保留本文链接:https://go.coder-hub.com/63040638.html
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