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leetcode/problems/src/array/MergeIntervals.java at master · KindleBooks66/leetcode · GitHub
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array
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MergeIntervals.java
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array
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MergeIntervals.java
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package
array
;
import
java
.
util
.
ArrayList
;
import
java
.
util
.
Arrays
;
import
java
.
util
.
Collections
;
import
java
.
util
.
List
;
/**
* Created by gouthamvidyapradhan on 13/06/2017. Given a collection of intervals, merge all
* overlapping intervals.
*
* <p>For example, Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18].
*
* <p>Solution: O(N log N) where N is the number of intervals 1. Sort the intervals based on start
* index 2. Mark the first interval as the current interval 3. For every ith interval starting 1 ->
* N, if the ith interval overlaps with the current interval then create a new current interval.
* Else, add the current interval to result set and begin a new current interval.
*/
public
class
MergeIntervals
{
public
static
class
Interval
{
int
start
;
int
end
;
Interval
() {
start
=
0
;
end
=
0
;
}
Interval
(
int
s
,
int
e
) {
start
=
s
;
end
=
e
;
}
}
public
static
void
main
(
String
[]
args
)
throws
Exception
{
Interval
i1
=
new
Interval
(
1
,
2
);
Interval
i2
=
new
Interval
(
3
,
4
);
Interval
i3
=
new
Interval
(
5
,
6
);
Interval
i4
=
new
Interval
(
1
,
10
);
List
<
Interval
>
result
=
new
MergeIntervals
().
merge
(
Arrays
.
asList
(
i1
,
i2
,
i3
,
i4
));
result
.
forEach
((
I
) ->
System
.
out
.
println
(
I
.
start
+
" "
+
I
.
end
));
}
public
List
<
Interval
>
merge
(
List
<
Interval
>
intervals
) {
if
(
intervals
.
isEmpty
())
return
new
ArrayList
<>();
Collections
.
sort
(
intervals
, (
o1
,
o2
) ->
Integer
.
compare
(
o1
.
start
,
o2
.
start
));
List
<
Interval
>
result
=
new
ArrayList
<>();
Interval
curr
=
intervals
.
get
(
0
);
for
(
int
i
=
1
,
l
=
intervals
.
size
();
i
<
l
;
i
++) {
Interval
I
=
intervals
.
get
(
i
);
if
(
I
.
start
>=
curr
.
start
&&
I
.
start
<=
curr
.
end
) {
// check if the new interval overlaps with the current
curr
.
end
=
curr
.
end
>
I
.
end
?
curr
.
end
:
I
.
end
;
}
else
{
result
.
add
(
curr
);
curr
=
I
;
}
}
result
.
add
(
curr
);
return
result
;
}
}
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