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:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
感觉要解这种题，首先要迅速对可能出现的情况进行分类，比如这道题可以分三类：

聪明的分类,对吧。然后中心思想是：

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，这意味着永远都和这个interval无关，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，和上面的原因类似，并更新newInterval为当前的interval，因为之后只有大的interval才能用上。 

当当前的interval与newInterval有重叠时，merge interval并更新新的newInterval为merge后的。大大，小小

/**
 * 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; }
 * }
 */
class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> res= new ArrayList<Interval>();
        int l = intervals.size();
        for(Interval each : intervals){
            if(each.end < newInterval.start){
                res.add(each);
            }
            else if(newInterval.end < each.start){
                res.add(newInterval);
                newInterval = each;
            }
            else if(newInterval.start <= each.end || each.start <= newInterval.end){
               newInterval.start = (Math.min(newInterval.start, each.start));
                newInterval.end = (Math.max(newInterval.end, each.end));
            }
        }
        res.add(newInterval);
        return res;
    }
}

挖个坑，后面有时间（福来阁）看看二分法怎么做。