Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

复用 Insert Interval代码即可。

 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 public class Solution {

11     public ArrayList<Interval> merge(ArrayList<Interval> intervals) {

12         // Note: The Solution object is instantiated only once and is reused by each test case.

13         ArrayList<Interval> results = new ArrayList<Interval>();

14         if(intervals == null || intervals.size() == 0) return intervals;

15         for(int i = 0; i < intervals.size(); i ++)

16         {

17             results = insert(results,intervals.get(i));

18         }

19         return results;

20     }

21 

22     public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {

23         // Note: The Solution object is instantiated only once and is reused by each test case.

24         ArrayList<Interval> results = new ArrayList<Interval>();

25         boolean sig = false;

26         for(int i = 0; i < intervals.size(); i ++)

27         {

28             if(intervals.get(i).start > newInterval.end)

29             {

30                 if(!sig)

31                 {

32                     results.add(newInterval);

33                     sig = true;

34                 }

35                 results.add(intervals.get(i));

36             }

37             else

38             {

39                 newInterval.end = Math.max(newInterval.end, intervals.get(i).end);

40             }

41             if(intervals.get(i).end < newInterval.start)

42             {

43                 results.add(intervals.get(i));    

44             }

45             else

46             {

47                 newInterval.start = Math.min(newInterval.start, intervals.get(i).start);

48             }

49         }

50          if(!sig)

51         {

52             results.add(newInterval);

53         }

54         return results;

55     }

56 }

 第二遍：

 1 public class Solution {

 2     public ArrayList<Interval> merge(ArrayList<Interval> intervals) {

 3         // Note: The Solution object is instantiated only once and is reused by each test case.

 4         ArrayList<Interval> result = new ArrayList<Interval>();

 5         if(intervals == null || intervals.size() == 0)return result;

 6         PriorityQueue<Interval> pq = new PriorityQueue<Interval>(intervals.size(), new Comparator<Interval>(){

 7             public int compare(Interval a, Interval b){

 8                 return a.start > b.start ? 1 : (a.start == b.start ? 0 : -1);

 9             }

10         });

11         for(Interval interval:intervals){

12               pq.add(interval);

13         }

14         result.add(pq.poll());//poll out the minimum element in the queue

15         int max_end = result.get(0).end;

16         while(pq.size() != 0){

17             Interval tmp = pq.poll();

18             if(tmp.start <= max_end){

19                 max_end = max_end > tmp.end ? max_end : tmp.end;

20                 result.get(result.size() - 1).end = max_end;

21             }else{

22                 result.add(tmp);

23                 max_end = tmp.end;

24             }

25         }

26         return result;

27     }

28 }

先把interval 按start time 排序，再添加。 保存当前的最大end time。