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].

思路：56 Merge Interval 基础上加一行代码

/**
 * 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) {
        intervals.add(newInterval);
        if (intervals.size()<=1){
            return intervals;
        }

        Comparator<Interval> comp = new Comparator<Interval>(){
            public int compare(Interval a, Interval b){
                return a.start - b.start;
            }
        };

        Collections.sort(intervals,comp);

        List<Interval> res = new ArrayList<Interval>();
        int start = intervals.get(0).start;
        int end = intervals.get(0).end;

        for (Interval inter:intervals){
            if (inter.start<=end){
                end = Math.max(end,inter.end);
            }else{
                res.add(new Interval(start,end));
                start = inter.start;
                end = inter.end;
            }
        }

        res.add(new Interval(start,end));
        return res;
    }
}

思路2：单纯比比比

    List<Interval> result = new ArrayList<Interval>();
    for (Interval i : intervals) {
        if (newInterval == null || i.end < newInterval.start)
            result.add(i);
        else if (i.start > newInterval.end) {
            result.add(newInterval);
            result.add(i);
            newInterval = null;
        } else {
            newInterval.start = Math.min(newInterval.start, i.start);
            newInterval.end = Math.max(newInterval.end, i.end);
        }
    }
    if (newInterval != null)
        result.add(newInterval);
    return result;
}