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].
Solution：
思路：
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
/**
* 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;
// Sort by ascending starting point using an anonymous Comparator
intervals.sort((i1, i2) -> i1.start - i2.start);
List<Interval> result = new LinkedList<Interval>();
int start = intervals.get(0).start;
int end = intervals.get(0).end;
for (Interval interval : intervals) {
if (interval.start <= end) // Overlapping intervals, move the end if needed
end = Math.max(end, interval.end);
else { // Disjoint intervals, add the previous one and reset bounds
result.add(new Interval(start, end));
start = interval.start;
end = interval.end;
}
}
// Add the last interval
result.add(new Interval(start, end));
return result;
}
}
