【Java】删除集合元素的正确与错误做法

我会带着你远行 2024-03-17 17:31 70阅读 0赞

## 错误做法 ##

### 一、fori正序 + list.remove(num) ###

@Test  
    public void test031(){
          
        ArrayList<Integer> list = new ArrayList<>();  
        list.add(1);  
        list.add(3);  
        list.add(3);  
        for (int i = 0; i < list.size(); i++) {
          
            Integer num=list.get(i);  
            if(num==3){
          
                list.remove(num);  
            }  
        }  
        list.forEach(System.out::println);// 1 3  
    }

![image.png][]

我们可以看到，集合中有两个3，结果只删掉了第一个，原因是删除第一个3之后，第二个3的下标变成1，缺少了一次检查  
![image.png][image.png 1]

改成方法：删除之后i–,回退一次，见下面的正确做法一

### 二、增强for + list.remove(num) ###

@Test  
    public void test07(){
          
        ArrayList<Integer> list = new ArrayList<>();  
        list.add(1);  
        list.add(3);  
        list.add(2);  
      
        for (Integer num:list) {
          
            if(num==1){
          
                list.remove(num);  
            }  
        }  
        list.forEach(System.out::println);  
    }

![image.png][image.png 2]

我们看到报错的是ArrayList内部迭代器的next方法中的checkForCommdification方法抛出ConcurrentModificationException,源码如下

public E next() {
          
        checkForComodification();  
        int i = cursor;  
        if (i >= size)  
            throw new NoSuchElementException();  
        Object[] elementData = ArrayList.this.elementData;  
        if (i >= elementData.length)  
            throw new ConcurrentModificationException();  
        cursor = i + 1;  
        return (E) elementData[lastRet = i];  
    }
    final void checkForComodification() {
         
    	
        if (modCount != expectedModCount)  
            throw new ConcurrentModificationException();  
    }

我们可以看到modCount != expectedModCount，而抛出异常  
原因是remove的时候修改了modCount,但是并未修改expectedModeCount

ArrayList的remove(Object o)的源码如下：

public boolean remove(Object o) {
          
        if (o == null) {
          
            for (int index = 0; index < size; index++)  
                if (elementData[index] == null) {
          
                    fastRemove(index);  
                    return true;            }  
        } else {
          
            for (int index = 0; index < size; index++)  
                if (o.equals(elementData[index])) {
          
                    fastRemove(index);  
                    return true;            }  
        }  
        return false;  
    }  
      
    /*  
     * Private remove method that skips bounds checking and does not * return the value removed. */
    private void fastRemove(int index) {
          
        modCount++;  
        int numMoved = size - index - 1;  
        if (numMoved > 0)  
            System.arraycopy(elementData, index+1, elementData, index,  
                             numMoved);  
        elementData[--size] = null; // clear to let GC do its work  
    }

修正方法：使用iteritor的remove(),见下面正确方法的三

## 正确做法 ##

### 一、fori正序 + list.remove(num) + 回退 ###

@Test  
    public void test03(){
          
        ArrayList<Integer> list = new ArrayList<>();  
        list.add(1);  
        list.add(3);  
        list.add(3);  
        for (int i = 0; i < list.size(); i++) {
          
            Integer num=list.get(i);  
            if(num==3){
          
                list.remove(num);  
                i--//注意回退，不然会缺少一次检查
            }  
        }  
        list.forEach(System.out::println);// 1 3  
    }

### 二、fori倒序 + list.remove(num) ###

@Test  
    public void test04(){
          
        ArrayList<Integer> list = new ArrayList<>();  
        list.add(1);  
        list.add(3);  
        list.add(2);  
        for (int i = list.size()-1; i >=0; i--) {
          
            Integer num=list.get(i);  
            if(num==1){
          
                list.remove(num);  
            }  
        }  
        list.forEach(System.out::println);  
    }

倒叙删除不用回退，如下图  
![image.png][image.png 3]

### 三、iterator.remove() ###

@Test  
    public void test01(){
          
        ArrayList<Integer> list = new ArrayList<>();  
        list.add(1);  
        list.add(3);  
        list.add(2);  
        Iterator<Integer> iterator = list.iterator();  
        while (iterator.hasNext()){
          
            Integer next = iterator.next();  
            if(next==3){
          
                iterator.remove();  
            }  
        }  
        list.forEach(System.out::println);  
    }

这种做法不会抛出ConcurrentModificationException的原因是，iterator.remove()也调用了remove(Object o),但是它把modCount的值，也赋值给了exceptionCount  
源码如下：

public void remove() {
          
        if (lastRet < 0)  
            throw new IllegalStateException();  
        checkForComodification();  
      
        try {
          
            ArrayList.this.remove(lastRet);  
            cursor = lastRet;  
            lastRet = -1;  
            expectedModCount = modCount;  //这里
        } catch (IndexOutOfBoundsException ex) {
          
            throw new ConcurrentModificationException();  
        }  
    }

### 四、list.removeIf(条件) ###

@Test  
    public void test02(){
          
        ArrayList<Integer> list = new ArrayList<>();  
        list.add(1);  
        list.add(3);  
        list.add(3);  
        list.removeIf(next -> next == 3);  
        list.forEach(System.out::println);  
    }

## 注意事项 ##

*  remove有两个重载方法，一个接收int的index,一个接收Object的o,注意不要写错

[image.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/15/ddb40e16b722491d85e7d97aadc09e36.png
[image.png 1]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/15/338d6b6d033844ab90467618da3bff2b.png
[image.png 2]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/15/0185832f8af14bb3ba9d4b577b63b55f.png
[image.png 3]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/15/ecd4ae082cc14c669abc75c4f71c7e44.png