352. Data Stream as Disjoint Intervals

NO IMAGE

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併。

/**
* 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<Integer, Interval> 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<Interval> getIntervals() {
return new ArrayList<Interval>(map.values());
}
}
/**
* Your SummaryRanges object will be instantiated and called as such:
* SummaryRanges obj = new SummaryRanges();
* obj.addNum(val);
* List<Interval> param_2 = obj.getIntervals();
*/