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].

解题思路为找到newInterval可以merge的start和end点。(start inclusive, end exclusive)

• start: interval.end 刚刚大于或等于 newInterval.start
• end: interval.start 刚刚大于 newInterval.end

/**
 * 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; }
 * }
 */
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> res = new ArrayList<>();
        if(intervals == null || intervals.size() == 0) {
            res.add(newInterval);
            return res;
        }
        int start = 0, end = 0;
        for(Interval i : intervals) {
            if(i.end < newInterval.start) start++;
            if(i.start <= newInterval.end) end++;
        }
        if(start == end) {
            res.addAll(intervals);
            res.add(start, newInterval);
            return res;
        }
        for(int i = 0; i < start; i++) res.add(intervals.get(i));
        Interval interval = new Interval(Math.min(intervals.get(start).start, newInterval.start), Math.max(intervals.get(end-1).end, newInterval.end));
        res.add(interval);
        for(int i = end; i < intervals.size(); i++) res.add(intervals.get(i));
        return res;
    }
}