Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
tag: 对象排序, List转换成数组
/** * 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> insert(List<Interval> intervals, Interval newInterval) { List<Interval> result = new ArrayList<Interval>(); intervals.add(newInterval); if(intervals == null || intervals.size() == 0) { return result; } int size = intervals.size(); Interval[] intervalArr = intervals.toArray(new Interval[size]); Arrays.sort(intervalArr, new Comparator<Interval>(){ public int compare(Interval interval1, Interval interval2) { return interval1.start - interval2.start; } }); Interval mover = intervalArr[0]; for(int i = 1; i < size; i++) { if(mover.end < intervalArr[i].start) { result.add(mover); mover = intervalArr[i]; } else { mover.end = Math.max(mover.end, intervalArr[i].end); } } result.add(mover); return result; } }