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].

 与merge intervals(合并间隔)有点类似。

1、newInterval与当前interval没有交集，则按照先后次序加入newInterval和当前interval，然后装入所有后续interval。返回ret。

2、newInterval与当前interval有交集，合并成为新的newInterval，然后处理后续interval。

3、处理完最后一个interval若仍未返回ret，说明newInterval为最后一个interval，装入ret。返回ret

/**
 * 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; }
 * }
 */
class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> res=new ArrayList<Interval>();
        
        int i=0;
        while(i<intervals.size()&&intervals.get(i).end<newInterval.start){
            res.add(intervals.get(i));
            i++;
        }
        //一个个比较，插入的与当前的合并成新的间隔插入 ，
        while(i<intervals.size()&&intervals.get(i).start<=newInterval.end){
            newInterval=new Interval(
                        Math.min(newInterval.start,intervals.get(i).start),
                        Math.max(newInterval.end,intervals.get(i).end)
                            );
            i++;
        }
        res.add(newInterval);
        while(i<intervals.size()){
            res.add(intervals.get(i));
            i++;
        }
        return res;
    }
}