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]
.
老娘提交了20次终于通过了的版本……
public class Solution { public List<Interval> merge(List<Interval> intervals) { if (intervals == null || intervals.size() <= 1) { return intervals; } Collections.sort(intervals, new IntervalComparator()); List<Interval> res = new ArrayList<Interval>(); int i=0; while(i<intervals.size()) { Interval newint = new Interval(); int newstart = intervals.get(i).start; int newend = intervals.get(i).end; int j=i+1; while(j<intervals.size()&&newend>=intervals.get(j).start) { newend = Math.max(newend,intervals.get(j).end); j++; } newint.start=newstart; newint.end=newend; res.add(newint); if(j!=i+1) i=j; else i++; } return res; } private class IntervalComparator implements Comparator<Interval> { public int compare(Interval a, Interval b) { return a.start - b.start; } } }