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 List<Interval> insert(List<Interval> intervals, Interval newInterval) {
    	List<Interval> res = new ArrayList<Interval>();
    	for (Interval in : intervals) {
    		if (newInterval.start > in.end) {
    			res.add(in);
    		} else if (newInterval.end < in.start) {
    			res.add(newInterval);
    			newInterval = in;
    		} else if (newInterval.start<=in.end || newInterval.end>=in.start) {
    			int start = Math.min(newInterval.start, in.start);
    			int end = Math.max(in.end, newInterval.end);
    			newInterval = new Interval(start, end);
    		}
    	}
    	res.add(newInterval);
    	return res;
    }
}

?

Insert Interval

原文：http://hcx2013.iteye.com/blog/2221111