352. Data Stream as Disjoint Interval

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?

本题考查TreeMap，这里区别一下。TreeMap可以检查某一个key值和它相临的比它大一点和比它小一点的key值，还可以用containsKey来检查是否包含该元素。检查的方法分别是lowerKey和higherKey。而TreeSet里面有ceiling和flooring两种方法，分别表示大于等于它和小于等于它的值。本题的key值为Interval里面的start值，value值为Interval。然后进行详细的判断即可。

/**

* 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<Integer,Interval>();

}

public void addNum(int val) {

if(map.containsKey(val)) return;

Integer lo = map.lowerKey(val);

Integer hi = map.higherKey(val);

if(lo!=null&&hi!=null&&map.get(lo).end+1==val&&hi==val+1){

map.get(lo).end = map.get(hi).end;

map.remove(hi);

}else if(lo!=null&&map.get(lo).end+1>=val){

map.get(lo).end = Math.max(val,map.get(lo).end);

}else if(hi!=null&&hi==val+1){

map.put(val,new Interval(val,map.get(hi).end));

map.remove(hi);

}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();

*/