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

这道题可以在Merge Intervals基础上完成，时间复杂度有点高。

/**
 * 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) {
        int low = newInterval.start;
        int high = newInterval.end;
        ListIterator<Interval> iterator = intervals.listIterator();
        
        while(iterator.hasNext()){
            Interval interval = iterator.next();
            if(high < interval.start){
                iterator.previous();
                iterator.add(new Interval(low, high));
                return intervals;
            }
            if(low > interval.end)
                continue;
            else{
                low = Math.min(low, interval.start);
                high = Math.max(high, interval.end);
                iterator.remove();
            }
        }//while
        intervals.add(new Interval(low, high));
        
        return intervals;
    }    
}