My awesome solution

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    public class Solution {
        public List<Interval> merge(List<Interval> intervals) {
            Collections.sort(intervals, c);
            Iterator<Interval> it = intervals.iterator();  
            Interval curr = null;
                 curr =;
                 Interval tmp =;
                 if(tmp.start <= curr.end){            
                     curr.end = Math.max(tmp.end, curr.end);
                    curr = tmp;
             return intervals;       
        private static final Comparator<Interval> c = new Comparator<Interval>(){
           public int compare(Interval a, Interval b)
               if(a.start != b.start)
                   return a.start - b.start;
                   return a.end - b.end;      

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    Since you already sorted the intervals, you don't have to do a "list remove" for each merge. Do X merges, and remove X elements at once. That can save a lot of time because remove operation really is O(N).

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    Thank for your suggestion. But I am not sure how removing X elements at once would save a lot of time. If you are talking about having a to-be-deleted list of intervals, and call removeAll method, then it wouldn't going to save time because ...

    public boolean removeAll(Collection<?> c) {
        boolean modified = false;
        Iterator<?> e = iterator();
        while (e.hasNext()) {
            if (c.contains( {
                modified = true;
        return modified;

    Otherwise, please let me know what do you have in mind.

  • 0

    After sorting, an Interval only can merge with its neighbors. Say, Interval[i] can merge with Interval[i+1...i+j], then after we do j merges (but not vector.erase()), we use one vector.erase() to remove Interval[i+1...i+j].

    Why? vector.erase() has a complexity of O(N). I am not sure if my solution runs faster than yours on OJ. Maybe it is just a small improvement. My solution runs in 612 ms (looks like this question has some big test case).

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