ArrayList 的 removeif

首先用了函数式的接口，作为过滤器。
remove 时需要考虑删除是 内存移位等问题

removeif 是这样解决的

1. 使用与list元素个数等长的 BitSet
2. 使用过滤器，过滤出需要删除的元素，同时在 bitSet 相应的位打上1
3. 此时知道没被删除的 元素个数 notDelNum
4. 从0开始循环， 将list元素的基于位图为 0的元素位置 重新排列位置
5. 将大于notDelNum的元素全部去掉

·

public boolean removeIf(Predicate<? super E> filter) {
        Objects.requireNonNull(filter);
        // figure out which elements are to be removed
        // any exception thrown from the filter predicate at this stage
        // will leave the collection unmodified
        int removeCount = 0;
        final BitSet removeSet = new BitSet(size);
        final int expectedModCount = modCount;
        final int size = this.size;
        for (int i=0; modCount == expectedModCount && i < size; i++) {
            @SuppressWarnings("unchecked")
            final E element = (E) elementData[i];
            if (filter.test(element)) {
                removeSet.set(i);
                removeCount++;
            }
        }
        if (modCount != expectedModCount) {
            throw new ConcurrentModificationException();
        }
        // shift surviving elements left over the spaces left by removed elements
        final boolean anyToRemove = removeCount > 0;
        if (anyToRemove) {
            final int newSize = size - removeCount;
            for (int i=0, j=0; (i < size) && (j < newSize); i++, j++) {
                i = removeSet.nextClearBit(i);
                elementData[j] = elementData[i];
            }
            for (int k=newSize; k < size; k++) {
                elementData[k] = null;  // Let gc do its work
            }
            this.size = newSize;
            if (modCount != expectedModCount) {
                throw new ConcurrentModificationException();
            }
            modCount++;
        }
        return anyToRemove;
    }