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].

思路

先按照Interval.start进行升序排列，如果前一个front,后一个behind。如果behind.end < front.start不用管，behind.end >= front.end 删除front。behind.end > front.start && behind.end < front.end合并

/**
 * 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) {
        Collections.sort(intervals, new Compare());
        boolean isMerge = false;
        while(true){
            isMerge = false;
            for(int i = 0; i < intervals.size() - 1; i++){
                Interval behind = intervals.get(i);
                Interval front    = intervals.get(i + 1);
                if(behind.end < front.start)                                    //后面一个end小于前面的start
                    continue;
                else if(behind.end >= front.end)                                //后面一个end大于等于前面的end
                {
                    intervals.remove(i + 1);
                    isMerge = true;
                }
                else if(behind.end < front.end && behind.end >= front.start){    //后面的end在前面的start和end中间，合并
                    behind.end = front.end;
                    intervals.remove(i + 1);
                    isMerge = true;
                }//else if
            }//for
            if(!isMerge)
                break;
        }
        
        return intervals;
    }

}
class Compare implements Comparator<Interval>{

    @Override
    public int compare(Interval o1, Interval o2) {
        if(o1.start < o2.start)
            return -1;
        else if(o1.start > o2.start)
            return 1;
        else
            return 0;
    }
}