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].

Solution：Interative

思路：

Screen Shot 2017-11-21 at 12.56.17.png

Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> result = new ArrayList<>();
        
        int i = 0;
        // add all the intervals ending before newInterval starts
        while(i < intervals.size() && newInterval.start > intervals.get(i).end) {
            result.add(intervals.get(i));
            i++;
        }
        
        int new_start = newInterval.start;
        int new_end = newInterval.end;
        // merge all overlapping intervals to one considering newInterval
        while(i < intervals.size() && newInterval.end >= intervals.get(i).start) {
            new_start = Math.min(new_start, intervals.get(i).start);
            new_end = Math.max(new_end, intervals.get(i).end);
            i++;
        }
        result.add(new Interval(new_start, new_end)); // add the union of intervals we got
        
        // add all the rest
        while (i < intervals.size()) {
            result.add(intervals.get(i));
            i++;
        }
        return result;
    }
}