# 57 Insert Interval

### Problem:

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].

### Solutions:

``````/**
* 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) {
boolean inserted = false;
for (Interval inter:intervals) {
if (inserted) {
}
else if (newInterval.start > inter.end) {
//no overlap
}
else if (newInterval.end < inter.start) {
inserted = true;
}
else {
newInterval.start = Math.min(newInterval.start, inter.start);
newInterval.end = Math.max(newInterval.end, inter.end);
}
}
if (!inserted) {