forked from mengli/leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
InsertInterval.java
56 lines (51 loc) · 1.44 KB
/
InsertInterval.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
/**
* Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
*
* You may assume that the intervals were initially sorted according to their start times.
*
* Example 1:
* Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
*
* Example 2:
* Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
*
* This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
*/
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
public class InsertInterval {
public class IntervalCmp implements Comparator<Interval> {
@Override
public int compare(Interval i1, Interval i2) {
if (i1.start < i2.start) {
return -1;
}
if (i1.start == i2.start && i1.end <= i2.end) {
return -1;
}
return 1;
}
}
public ArrayList<Interval> insert(ArrayList<Interval> intervals,
Interval newInterval) {
intervals.add(newInterval);
Interval[] arr = new Interval[intervals.size()];
intervals.toArray(arr);
Arrays.sort(arr, new IntervalCmp());
intervals.clear();
int start = arr[0].start;
int end = arr[0].end;
for (int i = 1; i < arr.length; i++) {
if (arr[i].start <= end) {
end = Math.max(end, arr[i].end);
} else {
intervals.add(new Interval(start, end));
start = arr[i].start;
end = arr[i].end;
}
}
intervals.add(new Interval(start, end));
return intervals;
}
}