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:
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].

本题与上一题解法相同，只需要在原来列表中加入insert的元素，再做和56题相同的操作即可，程序如下：

/** * 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) {        if (intervals.size() == 0){            List<Interval> l = new ArrayList<>();            l.add(newInterval);            return l;        }        intervals.add(newInterval);        List<Interval> lst = new ArrayList<>();        Collections.sort(intervals, new Mycomparator());        lst.add(intervals.get(0));        int size = intervals.size();        for (int i = 1; i < size; ++ i){            Interval top = lst.get(lst.size() - 1);            if (top.end < intervals.get(i).start){                lst.add(intervals.get(i));            }            else {                Interval tmp = new Interval(Math.min(top.start, intervals.get(i).start),                                             Math.max(top.end, intervals.get(i).end));                lst.remove(lst.size() - 1);                lst.add(tmp);            }        }        return lst;    }}class Mycomparator implements Comparator<Interval> {    @Override    public int compare(Interval l0, Interval l1){        return l0.start - l1.start;    }}

当然还有另外一种更高效率的做法，因为题目已经说明，原来的list是根据start有序排列的，所以，只需扫描list中Interval对象，保留与insert不重叠的Interval对象，更新重叠的区域，相对第一种解法省去了sort的过程。