352. Data Stream as Disjoint Intervals

Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals.

For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be:
[1, 1]
[1, 1], [3, 3]
[1, 1], [3, 3], [7, 7]
[1, 3], [7, 7]
[1, 3], [6, 7]

Follow up:
What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?

Solution：TreeMap

思路： 不能用bucket形式保存，因为范围未知，所以只能用直接保存值的形式。为了to easily find the lower and higher keys( for merge purpose)( key is the start of the interval)，所以利用TreeMap。具体通过higherKey, lowerKey API
Time Complexity: addNum: O(1) getIntervals: O(logN)
Space Complexity: O(N)

Solution Code：

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class SummaryRanges {
    TreeMap tree;

    public SummaryRanges() {
        tree = new TreeMap<>();
    }

    public void addNum(int val) {
        if(tree.containsKey(val)) return;
        Integer l = tree.lowerKey(val);
        Integer h = tree.higherKey(val);
        
        if(l != null && h != null && tree.get(l).end + 1 == val && h == val + 1) {
            // merge with low and high
            tree.get(l).end = tree.get(h).end;
            tree.remove(h);
        } else if(l != null && tree.get(l).end + 1 == val) {
            // merge with low
            tree.get(l).end = val;
        } else if(h != null && h == val + 1) {
            // merge with high
            tree.put(val, new Interval(val, tree.get(h).end));
            tree.remove(h);
        } else if(l != null && tree.get(l).end + 1 > val) {
            // already covered, then do nothing
            ;
        }
        else {
            // insert new
            tree.put(val, new Interval(val, val));
        }
    }

    public List getIntervals() {
        return new ArrayList<>(tree.values());
    }
}