Java HashMap在遍历时删除元素的实现

import java.util.HashMap;
import java.util.Map;

public class MapEntryExample {
    public static void main(String[] args) {
        Map<Integer, String> maps = new HashMap<>();
        maps.put(3, "张飞");
        maps.put(2, "关羽");
        maps.put(4, "赵云");
        maps.put(1, "刘备");
        maps.put(5, "马超");

        for (Map.Entry<Integer, String> entry : maps.entrySet()) {
            System.out.println("key=" + entry.getKey() + " value=" + entry.getValue());
        }
    }
}
这段代码会输出Map中每个键值对的键和值：

key=1 value=刘备
key=2 value=关羽
key=3 value=张飞
key=4 value=赵云
key=5 value=马超
通过使用Map.Entry，你可以方便地遍历Map集合并获取每个键值对的键和值。

    public static void main(String[] args) {
        Map<Integer, String> stuMap = new HashMap<>();
        stuMap.put(1, "xlsj");
        stuMap.put(2, "xinliushijian");
        stuMap.put(3, "心流时间");

        // A: entrySet()、keySet()、values()方法作用
        // key:value 集合
        Set<Map.Entry<Integer, String>> stuEntrySet = stuMap.entrySet();
        // key 集合
        Set<Integer> stuKeySet = stuMap.keySet();
        // value 集合
        Collection<String> stuValues = stuMap.values();

        System.out.println(stuEntrySet); // [1=xlsj, 2=xinliushijian, 3=心流时间]
        System.out.println(stuKeySet); // [1, 2, 3]
        System.out.println(stuValues); // [xlsj, xinliushijian, 心流时间]


        // B: 遍历集合时删除元素, 普通for循环、迭代器不会报错，其他会报错，增强for循环其实也是用到迭代器,
        // 集合遍历时不可以删除元素，而迭代器可以
        // 报错原因如下modCount != expectedModCount
/*        final HashMap.Node<K,V> nextNode() {
            HashMap.Node<K,V>[] t;
            HashMap.Node<K,V> e = next;
            if (modCount != expectedModCount)
                throw new ConcurrentModificationException();
            if (e == null)
                throw new NoSuchElementException();
            if ((next = (current = e).next) == null && (t = table) != null) {
                do {} while (index < t.length && (next = t[index++]) == null);
            }
            return e;
        }*/
        

        // 1.普通for循环遍历时可以删除元素
        for (int i = 0; i < stuKeySet.size(); i++) {
            if (i == 2) {
                stuKeySet.remove(i);
            }
        }

        // 2.for增强循环遍历时删除元素报异常：Exception in thread "main" java.util.ConcurrentModificationException
        for (int i : stuKeySet) {
            if (i == 2) {
                stuKeySet.remove(i);
            }
        }

        // 3.forEach循环遍历时删除元素报异常：Exception in thread "main" java.util.ConcurrentModificationException
        stuMap.forEach((k, v) -> {
            if (k == 2) {
                stuKeySet.remove(k);
            }
        });

        // 4.stream forEach循环遍历时删除元素报异常：Exception in thread "main" java.util.ConcurrentModificationException
        stuKeySet.stream().forEach(k -> {
            if (k == 2) {
                stuKeySet.remove(k);
            }
        });

        // 5.迭代器遍历时可以删除元素
        Iterator integers = stuEntrySet.iterator();
        while (integers.hasNext()) {
            Map.Entry<Integer, String> entry = (Map.Entry<Integer, String>) integers.next();
            System.out.println(entry.getKey());
            if (entry.getKey() == 2) {
                integers.remove();
            }

        }

        System.out.println(stuMap);
    }

    public static void main(String[] args) {
        Map<String, List<String>> map = new HashMap<>();
        List<String> list = new ArrayList<>();
        list.add("a");
        list.add("b");
        list.add("c");
        map.put("xin", list);
        map.put("xi", new ArrayList<>());
        map.put("x", null);

        // jing这个key不存在或者对应的value不存在，则put给的值
        List<String> list1 = map.computeIfAbsent("jing",h -> Arrays.asList(new String[]{"DE", "BU", "DE"}));
        List<String> list2 = map.computeIfAbsent("xin",h -> Arrays.asList(new String[]{"DE", "BU", "DE"}));
        List<String> list3 = map.computeIfAbsent("xi",h -> Arrays.asList(new String[]{"DE", "BU", "DE"}));
        List<String> list4 = map.computeIfAbsent("x",h -> Arrays.asList(new String[]{"DE", "BU", "DE"}));
        System.out.println(list1);
        System.out.println(list2);
        System.out.println(list3);
        System.out.println(list4);
        System.out.println("---------------jing--------------");
        System.out.println(map.get("jing"));

        map.putIfAbsent("jing1", Arrays.asList(new String[]{"DE1", "BU1", "DE1"}));
        System.out.println("---------------jing1--------------");
        System.out.println(map.get("jing1"));
    }