56. Merge Intervals

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].

Analysis:

Make a comparator, sort intervals by start.

iterate through all intervals, if overlap(previous end greater than interval.start), then keep updating temp end. till end < interval.start. Then add a new interval object to results.

Complexity:

Time: O(nlogn)

Space: O(n)

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; }
 * }
 */
public class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        List<Interval> returnList = new ArrayList<Interval>();
        if(intervals == null || intervals.size() == 0){
            return returnList;
        }
        Comparator intervalComparator = new Comparator<Interval>(){
            @Override
            public int compare(Interval a, Interval b){
                if(a.start == b.start){
                    return 0;
                }
                if(a.start < b.start){
                    return -1;
                }

                return 1;
            }
        };

        Collections.sort(intervals, intervalComparator);

        int start = intervals.get(0).start;
        int end = intervals.get(0).end;
        for(Interval itr : intervals){
            if(end >= itr.start){
                end = Math.max(end, itr.end);
            }
            else{
                returnList.add(new Interval(start, end));
                start = itr.start;
                end = itr.end;
            }
        }

        returnList.add(new Interval(start, end));
        return returnList;

    }
}