在不到O（n）的时间内获得数据结构中元素之间的最小差异

Question

Say I have some integers in a data structure. When I insert new number into the data structure, I want to get the smallest difference between the new inserted elements and any other elements already in the data structure. What data structure and algorithm should I use? AO(n) solution is trivial and I want better. Thanks.

Answer 1

One option would be to use a TreeSet (based on TreeMap ), which would require several O(lg n) operations. The class exposes two methods which can be used to find the element which is closest to the value you wish to insert:

public E ceiling(E e)
Returns the least element in this set greater than or equal to the given element, or null if there is no such element.

public E floor(E e)
Returns the greatest element in this set less than or equal to the given element, or null if there is no such element.

public static int findClosest(TreeSet set, Integer val) {
    if (set == null || set.size() == 0) {
        return -1;
    }

    // ceiling == 9 for input of 7
    // O(lg n) operation
    Integer ceiling = (Integer)set.ceiling(val);
    // floor = 6 for input of 7
    // O(lg n) operation
    Integer floor = (Integer)set.floor(val);

    if (ceiling == null) {
        return val - floor;
    }
    if (floor == null) {
        return ceiling - val;
    }

    return (val - floor > ceiling - val) ? ceiling - val : val - floor;
}

public static void main(String[] args) {
    TreeSet<Integer> ts = new TreeSet<>();
    ts.add(5);
    ts.add(1);
    ts.add(6);
    ts.add(9);
    ts.add(2);
    ts.add(3);

    int diff = findClosest(ts, 7);
    // closest is 6, so diff == 1
}

Answer 2

You could use a height-balanced binary search tree . These can be implemented with one of a number of data structures. Insertions are usually O(log(n)) on average and looking up the one or two closest integers (on either side of the inserted value) will also be at most O(log(n)).

There may be other data structures that can do better (particularly if you can reasonably bound the integer values you need to deal with), but I can't think of one off the top of my head.