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> added = new ArrayList<Interval>();
        boolean inserted = false;
        for(Interval inter: intervals) {
            if(!inserted && inter.start > newInterval.start) {
                added.add(newInterval);
                inserted = true;
            }
            added.add(inter);
        }
        if(!inserted) added.add(newInterval);

        Interval last = added.get(0);
        List<Interval> ans = new ArrayList<Interval>();
        for(int i = 1; i < added.size(); i++) {
            if(last.end >= added.get(i).start) {
                last.end = Integer.max(last.end, added.get(i).end);
            } else {
                ans.add(last);
                last = added.get(i);
            }
        }
        ans.add(last);
        return ans;
    }
}