/***
 * 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].
 */
/**
 * 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; }
 * }
 */
import java.util.List;

public class Solution {
    // Solution : O(n) O(1)
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        int i = -1;
        while(++i < intervals.size()) {
            if (newInterval.end < intervals.get(i).start) {
                intervals.add(i, newInterval);
                return intervals;
            } else if (newInterval.start > intervals.get(i).end) {
                continue;
            } else {
                newInterval.start = Math.min(intervals.get(i).start, newInterval.start);
                newInterval.end = Math.max(intervals.get(i).end, newInterval.end);
                intervals.remove(i--);
            }
        }
        intervals.add(intervals.size(), newInterval);
        return intervals;
    }
}

add InsertInterval.java