Java中TreeSet部分视图的size()复杂度是多少

Question

我想知道size()TreeSet的部分视图的时间复杂度是多少.

假设我正在添加随机数来设置(我不关心重复):

    final TreeSet<Integer> tree = new TreeSet<Integer>();
    final Random r = new Random();
    final int N = 1000;
    for ( int i = 0; i < N; i++ ) {
        tree.add( r.nextInt() );
    }

现在我正在考虑size()调用的复杂性:

    final int M = 100;
    for ( int i = 0; i < M; i++ ) {
        final int f = r.nextInt();
        final int t = r.nextInt();
        System.out.println( tree.headSet( t ).size() );
        System.out.println( tree.tailSet( f ).size() );
        if ( f > t ) {
            System.out.println( tree.subSet( t, f ).size() );
        } else {
            System.out.println( tree.subSet( f, t ).size() );
        }
    }

AFAIK的复杂性tree.headSet( t ),tree.tailSet( f )并且tree.subSet( f, t )是O(LG N),set.size()是O(1),但对于size()上述的方法？我有一种不好的感觉,它是O(K),其中K是所选子集的大小.

也许如果有一些解决方法来找到集合中某个元素的索引就足够了,因为如果我能得到ti = indexOf(f),那就说O(lg N)比它正是我需要的.

Answer 1

貌似复杂的size ()就是O(N),因为它可以调用TreeMap.NavigableSubMap.EntrySetView.size ()它像这样(甲骨文JDK 1.7.0_13)来实现:

public int size() {
    if (fromStart && toEnd)
        return m.size();
    if (size == -1 || sizeModCount != m.modCount) {
        sizeModCount = m.modCount;
        size = 0;
        Iterator i = iterator();
        while (i.hasNext()) {
            size++;
            i.next();
        }
    }
    return size;
}