原题链接在这里：https://leetcode.com/problems/data-stream-as-disjoint-intervals/description/
题目：
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?
题解：
利用TreeMap<Integert, Interval> tm来保存每段Interval的start 和 Interval本身的对应关系.
新值进来就看能不能连上已有的Interval.
Time Complexity: addNum, O(logn). getIntervals, O(n).
Space: O(n).
AC Java:
1 /** 2 * Definition for an interval. 3 * public class Interval { 4 * int start; 5 * int end; 6 * Interval() { start = 0; end = 0; } 7 * Interval(int s, int e) { start = s; end = e; } 8 * } 9 */ 10 class SummaryRanges { 11 TreeMap<Integer, Interval> tm; 12 /** Initialize your data structure here. */ 13 public SummaryRanges() { 14 tm = new TreeMap<Integer, Interval>(); 15 } 16 17 public void addNum(int val) { 18 if(tm.containsKey(val)){ 19 return; 20 } 21 Integer l = tm.lowerKey(val); 22 Integer r = tm.higherKey(val); 23 if(l!=null && r!=null && tm.get(l).end+1==val && val+1==r){ 24 tm.get(l).end = tm.get(r).end; 25 tm.remove(r); 26 }else if(l!=null && tm.get(l).end+1>=val){ 27 tm.get(l).end = Math.max(val, tm.get(l).end); 28 }else if(r!=null && val+1==r){ 29 tm.put(val, new Interval(val, tm.get(r).end)); 30 tm.remove(r); 31 }else{ 32 tm.put(val, new Interval(val, val)); 33 } 34 } 35 36 public List<Interval> getIntervals() { 37 return new ArrayList<Interval>(tm.values()); 38 } 39 } 40 41 /** 42 * Your SummaryRanges object will be instantiated and called as such: 43 * SummaryRanges obj = new SummaryRanges(); 44 * obj.addNum(val); 45 * List<Interval> param_2 = obj.getIntervals(); 46 */
