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].
Solution:
/**
* 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> ans = new ArrayList<>();
int i=0;
while(i<intervals.size() && intervals.get(i).end < newInterval.start) ans.add(intervals.get(i++));
int start = i<intervals.size() ? Math.min(intervals.get(i).start, newInterval.start) : newInterval.start;
while(i<intervals.size() && intervals.get(i).start <= newInterval.end) i++;
int end = i>0 ? Math.max(intervals.get(i-1).end, newInterval.end) : newInterval.end;
ans.add(new Interval(start, end));
while(i<intervals.size()) ans.add(intervals.get(i++));
return ans;
}
}