Given a collection of intervals, merge all overlapping intervals.

For example,
Given[1,3],[2,6],[8,10],[15,18],
return[1,6],[8,10],[15,18].

Solution: 先按start从小到大排序。然后比较cur.start与prev.end，如果prev.end < cur.start，说明不用合并，直接加入；否则把prev.end改为Math.max(prev.end, cur.end)

    public List<Interval> merge(List<Interval> intervals) {
        if (intervals == null || intervals.size() < 2) {
            return intervals;
        }
        Collections.sort(intervals, new Comparator<Interval>() {
            public int compare(Interval a, Interval b) {
                return a.start - b.start;
            }
        });
        List<Interval> ans = new ArrayList<>();
        for (Interval cur : intervals) {
            if (ans.size() == 0 || ans.get(ans.size() - 1).end < cur.start) {
                ans.add(cur);
            } else {
                ans.get(ans.size() - 1).end = Math.max(ans.get(ans.size() - 1).end, cur.end);
            }
        }
        return ans;
    }