LeetCode进阶之路（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:
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].

题目：在上一题的基础上，在加一个新的对象，也是合并的要求。

思路：直接加入到list里，用上一题的方法就可以了。

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        intervals.add(newInterval);
        if (intervals == null || intervals.size() <= 1) {
            return intervals;
        }
        
        Collections.sort(intervals, new IntervalComparator());       
  
        ArrayList<Interval> result = new ArrayList<Interval>();
        Interval last = intervals.get(0);
        for (int i = 1; i < intervals.size(); i++) {
            Interval curt = intervals.get(i);
            if (curt.start <= last.end ){
                last.end = Math.max(last.end, curt.end);
            }else{
                result.add(last);
                last = curt;
            }
        }
        
        result.add(last);
        return result;
    }
    private class IntervalComparator implements Comparator<Interval> {
        public int compare(Interval a, Interval b) {
            return a.start - b.start;
        }
    }

种一棵树最好的时间是十年前，其次是现在！