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].
思路
先按照Interval.start进行升序排列，如果前一个front,后一个behind。如果behind.end < front.start不用管，behind.end >= front.end 删除front。behind.end > front.start && behind.end < front.end合并
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 List<Interval> merge(List<Interval> intervals) {
12 Collections.sort(intervals, new Compare());
13 boolean isMerge = false;
14 while(true){
15 isMerge = false;
16 for(int i = 0; i < intervals.size() - 1; i++){
17 Interval behind = intervals.get(i);
18 Interval front = intervals.get(i + 1);
19 if(behind.end < front.start) //后面一个end小于前面的start
20 continue;
21 else if(behind.end >= front.end) //后面一个end大于等于前面的end
22 {
23 intervals.remove(i + 1);
24 isMerge = true;
25 }
26 else if(behind.end < front.end && behind.end >= front.start){ //后面的end在前面的start和end中间，合并
27 behind.end = front.end;
28 intervals.remove(i + 1);
29 isMerge = true;
30 }//else if
31 }//for
32 if(!isMerge)
33 break;
34 }
35
36 return intervals;
37 }
38
39 }
40 class Compare implements Comparator<Interval>{
41
42 @Override
43 public int compare(Interval o1, Interval o2) {
44 if(o1.start < o2.start)
45 return -1;
46 else if(o1.start > o2.start)
47 return 1;
48 else
49 return 0;
50 }
51 }
