/*

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; }

* }

*/

// Insert the interval first, then MergeInterval

// Note the case that newInterval is added to the end of the list

// time: O(n); space: O(1)

public class Solution {

public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {

if (intervals==null || intervals.size()==0){

intervals = new ArrayList<Interval>(); intervals.add(newInterval);

return intervals;

}

int i = 0;

while (i<intervals.size() && intervals.get(i).start < newInterval.start)

i++;

if (i==intervals.size()) intervals.add(newInterval); // Note here

else intervals.add(i, newInterval);

i = 0;

while (i<intervals.size()-1){

Interval curr = intervals.get(i), next = intervals.get(i+1);

if (curr.end<next.start)

i++;

else{

curr.start = Math.min(curr.start, next.start);

curr.end = Math.max(curr.end, next.end);

intervals.remove(i+1);

}

}

return intervals;

}

}