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  
}

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

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);  
}

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

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);  
}