Table of Contents
1.Map interface introduction
2.Map interface analysis
Home Java javaTutorial How to use Map in Java collection class

How to use Map in Java collection class

May 05, 2023 am 11:43 AM
java map

1.Map interface introduction

Map is used to save data with mapping relationships: Key - Value

For Set, the bottom layer is actually still a Map, but Set chooses not to use Value. That is to say, the Value value of Set is always a constant.

The Key and Value in Map can be any type of data and will be encapsulated into the HashMap$Node object.

The Key in Map cannot be repeated. But Value can be repeated. When there is the same Key, equivalence and replacement operations

2.Map interface analysis

The Map key-value pairs are stored in HashMap$Node

tab[i] = newNode(hash, key, value, null);

--

Node<K,V> newNode(int hash, K key, V value, Node<K,V> next) {
    return new Node<>(hash, key, value, next);
}
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So what is the data type of this Node node?

Let's take a look at the source code: (It also implements the Entry interface)

static class Node<K,V> implements Map.Entry<K,V> {
    final int hash;  // 存放hash值
    final K key;  // 存放key值
    V value;  // 存放Value值
    Node<K,V> next;  // 存放下一个个节点,以形成链表结构
}
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k-v In order to facilitate the programmer's traversal, an EntrySet collection will also be created. The element type stored in this collection is Entry, and an Entry object contains k, v, but essentially the k-v value here still points to the data in a Node node, that is, the k-v here still stores the address data

// k-v存在有EntrySet的一个指向
Set set = map.entrySet();
System.out.println(set.getClass());
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Output:

class java.util.HashMap$EntrySet

So, how is this feature that facilitates programmer traversal reflected?

The reason is that Map.Entry provides two very important methods: K getKey(); V getValue();. So we can traverse the Map in the following way:

Set set = map.entrySet();
System.out.println(set.getClass());
for (Object obj : set) {
    Map.Entry entry = (Map.Entry) obj;
    System.out.println(entry.getKey());
    System.out.println(entry.getValue());
}
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So how to prove that the k-v stored in Entry is just the pointer to the address? Very simple

Through debugging, let’s first take a look at the data stored in the Map. The address of the no2 key is @727

How to use Map in Java collection class

and then take a look at the set The address of the no2 key is exactly the same:

How to use Map in Java collection class

##3.Map interface method

Code example:

Map map = new HashMap();
// 添加键值对
map.put("no1","dahe");
map.put("no2","zhangsan");
// Key重复会进行替换
map.put("no1","lisi");
System.out.println(map);

// 根据键删除映射关系
map.remove("no1");
System.out.println(map);
// 根据key得到值
Object no2 = map.get("no2");
System.out.println(no2);
// 获取键值对数量
System.out.println(map.size());
// 判空
System.out.println(map.isEmpty());
// 清空
// map.clear();
// 查找键是否存在
System.out.println(map.containsKey("ok"));
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4.Map traversal method

The first formula: take out all the keys, and get the corresponding value through the key

// 取出所有的key,通过key取出对应的value
Set keySet = map.keySet();
for (Object o : keySet) {
    System.out.println(o);
    System.out.println(map.get(o));
}
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The second formula: take out all the value values

// 把所有的value值取出
Collection values = map.values();
for (Object value : values) {
    System.out.println(value);
}
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The third formula: through the EntrySet to get k-v

// 通过EntrySet来获取 k-v
Set entrySet1 = map.entrySet();
for (Object o : entrySet1) {
    // 将entry 转成 Map.Entry
    Map.Entry m = (Map.Entry) o;
    System.out.println(m.getKey());
    System.out.println(m.getValue());
}
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