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; }

* }

*/

public class Solution {

public ArrayList<Interval> insert(ArrayList<Interval> intervals,

Interval newInterval) {

Interval temp = newInterval;

ArrayList<Interval> lst = new ArrayList<Interval>();

for(Interval i : intervals) {

if(i.end < temp.start) {

lst.add(i);

} else if(i.start > temp.end){

lst.add(temp);

temp = i;

} else {

temp = new Interval(Math.min(temp.start, i.start),

Math.max(temp.end, i.end));

}

}

lst.add(temp);

return lst;

}

}