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();
*/
