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;
}
}