题目：
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].
题解：
这道题不仅要insert newInterval同时还要保证能够merge。那么就分情况讨论。
遍历每一个已给出的interval，
当当前的interval的end小于newInterval的start时，说明新的区间在当前遍历到的区间的后面，并且没有重叠，所以res添加当前的interval；
当当前的interval的start大于newInterval的end时，说明新的区间比当前遍历到的区间要前面，并且也没有重叠，所以把newInterval添加到res里，并更新newInterval为当前的interval；
当当前的interval与newInterval有重叠时，merge interval并更新新的newInterval为merge后的。
代码如下：
1 public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
2 ArrayList<Interval> res = new ArrayList<Interval>();
3
4 for(Interval each: intervals){
5 if(each.end < newInterval.start){
6 res.add(each);
7 }else if(each.start > newInterval.end){
8 res.add(newInterval);
9 newInterval = each;
}else if(each.end >= newInterval.start || each.start <= newInterval.end){
int nstart = Math.min(each.start, newInterval.start);
int nend = Math.max(newInterval.end, each.end);
newInterval = new Interval(nstart, nend);
}
}
res.add(newInterval);
return res;
}
Reference：//http://www.programcreek.com/2012/12/leetcode-insert-interval/