57. Insert Interval

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].

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

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> resu = new ArrayList();
        int s = newInterval[0], e = newInterval[1];
        int i = 0;
        //add all intervals which are before the new Interval
        while(i < intervals.length && intervals[i][1] < s){
            resu.add(intervals[i++]);
        }
        //ensure all of the intervals are merged based on the start and end time of the newInterval 
        while(i < intervals.length && intervals[i][0] <= e){
            s = Math.min(s, intervals[i][0]);
            e = Math.max(e, intervals[i][1]);
            i++;
        }
        //Add merged interval
        resu.add(new int[]{s, e});
        //Add the remaining intervals
        while(i < intervals.length) resu.add(intervals[i++]);
        int j = 0;
        int[][] res = new int[resu.size()][2];
        for(int[] tmp: resu){
            res[j++] = tmp;
        }
        return res;
    }
}