## 56. Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example 1:

``````Input:
[[1,3],[2,6],[8,10],[15,18]]

Output:
[[1,6],[8,10],[15,18]]

Explanation:
Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
``````

Example 2:

``````Input:
[[1,4],[4,5]]

Output:
[[1,5]]

Explanation:
Intervals [1,4] and [4,5] are considered overlapping.
``````

Thoughts:

1. Sort the interval list according the start point
2. record end, for loop check whether the current interval's start time is after the record end; if it is, merge the interval by maxing the end time; if it is not, insert the interval of recorded start and end time and update them to start another interval to be merged with current start and end time.

Code Java O(nlogn): Java8 lambda comparator

``````/**
* Definition for an interval.
* public class Interval {
*     int start;
*     int end;
*     Interval() { start = 0; end = 0; }
*     Interval(int s, int e) { start = s; end = e; }
* }
*/
class Solution {
public List<Interval> merge(List<Interval> intervals) {
if(intervals.size() <= 1) return intervals;

intervals.sort((i1, i2)->Integer.compare(i1.start, i2.start)); // java8 lambda

int start = intervals.get(0).start, end =intervals.get(0).end;
List<Interval> results = new LinkedList <Interval>();
for(Interval i: intervals){
if(i.start <= end){
end = Math.max(i.end, end);           // merging
}
else{
results.add(new Interval(start, end));   // disjoint: first adding existing merged intervals, then update start, end
start = i.start; end = i.end; // update start, end
}
}

return results;
}
}
``````

Code Python

``````# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
def merge(self, intervals):
"""
:type intervals: List[Interval]
:rtype: List[Interval]
"""
intervals =sorted(intervals, key=lambda x:x.start)
results = []

l = len(intervals)
i = 0
while i < l:
s = intervals[i].start
e = intervals[i].end
j = i + 1
while j < l and e >= intervals[j].start:
e = max(e, intervals[j].end)
j += 1
results.append(Interval(s,e))
i = j
return results
``````