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].

/**
 * 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 class IntervalComparator implements Comparator<Interval> {
    public int compare(Interval i0, Interval i1) {
        return i0.start - i1.start;
    }
}

public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
    ArrayList<Interval> newIntervals = new ArrayList<Interval>();
    newIntervals.addAll(intervals);
    newIntervals.add(newInterval);
    ArrayList<Interval> result = new ArrayList<Interval>();
    
    if ((intervals == null && newInterval == null)) {
        return result;
    }
    
    if (intervals == null || intervals.size() == 0) {
        result.add(newInterval);
        return result;
    }
    
    Collections.sort(newIntervals, new IntervalComparator());
    
    Interval last = newIntervals.get(0);
    for (int i = 1; i < newIntervals.size(); i++) {
        Interval curt = newIntervals.get(i);
        if (last.end >= curt.start) {
            last.end = Math.max(last.end, curt.end);
        } else {
            result.add(last);
            last = curt;
        }
    }
    result.add(last);
    return result;
}
}

思路和 merge interval 是几乎一样的。

就是把 newInterval 加入后，形成一个新的ArrayList，然后对其使用merge interval的方法