摘要：題目解答這題用會通過很快定位到當前數所處在哪兩個之間，從而進行高效的比較與合并。

題目：
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?

解答：
這題用Maptree會通過map.lowerKey, map.higherKey很快定位到當前數所處在哪兩個interval之間，從而進行高效的比較與合并。

/**
 * 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; }
 * }
 */
public class SummaryRanges {
    TreeMap map;
    /** Initialize your data structure here. */
    public SummaryRanges() {
        map = new TreeMap<>();
    }
    
    public void addNum(int val) {
        if (map.containsKey(val)) return;
        Integer l = map.lowerKey(val);
        Integer h = map.higherKey(val);
        if (l != null && h != null && map.get(l).end + 1 == val && val + 1 == map.get(h).start) {
            map.get(l).end = map.get(h).end;
            map.remove(h);
        } else if (l != null && val <= map.get(l).end + 1) {
            map.get(l).end = Math.max(map.get(l).end, val);
        } else if (h != null && map.get(h).start - 1 == val) {
            map.put(val, new Interval(val, map.get(h).end));
            map.remove(h);
        } else {
            map.put(val, new Interval(val, val));
        }
    }
    
    public List getIntervals() {
        return new ArrayList(map.values());
    }
}

/**
 * Your SummaryRanges object will be instantiated and called as such:
 * SummaryRanges obj = new SummaryRanges();
 * obj.addNum(val);
 * List param_2 = obj.getIntervals();
 */