Java中for循环内修改集合的常见陷阱与最佳实践

List<Integer> numbers = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
for (Integer num : numbers) {
    if (num % 2 == 0) {
        numbers.remove(num); // 抛出ConcurrentModificationException
    }
}

List<String> list = new ArrayList<>(Arrays.asList("A", "B", "C"));
for (String s : list) {
    if (s.equals("B")) {
        list.remove(s); // 抛出ConcurrentModificationException
    }
}

List<Integer> nums = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
for (int i = 0; i < nums.size(); i++) {
    if (nums.get(i) % 2 == 0) {
        nums.remove(i); // 可能导致元素跳过
    }
}
// 结果可能是 [1, 3, 4] 而非预期的 [1, 3]

final void checkForComodification() {
    if (modCount != expectedModCount)
        throw new ConcurrentModificationException();
}

List<Integer> numbers = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
Iterator<Integer> it = numbers.iterator();
while (it.hasNext()) {
    Integer num = it.next();
    if (num % 2 == 0) {
        it.remove(); // 安全删除
    }
}
System.out.println(numbers); // [1, 3]

List<Integer> numbers = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
numbers.removeIf(num -> num % 2 == 0);
System.out.println(numbers); // [1, 3]

List<Integer> numbers = new CopyOnWriteArrayList<>(Arrays.asList(1, 2, 3, 4));
for (Integer num : numbers) {
    if (num % 2 == 0) {
        numbers.remove(num); // 安全但性能较低
    }
}
System.out.println(numbers); // [1, 3]

List<Integer> numbers = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
for (int i = numbers.size() - 1; i >= 0; i--) {
    if (numbers.get(i) % 2 == 0) {
        numbers.remove(i); // 避免索引错位
    }
}
System.out.println(numbers); // [1, 3]

List<Integer> numbers = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
List<Integer> toRemove = new ArrayList<>();
for (Integer num : numbers) {
    if (num % 2 == 0) {
        toRemove.add(num);
    }
}
numbers.removeAll(toRemove);
System.out.println(numbers); // [1, 3]