[leedcode 56] 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) {
        //遍历intervals，并和newInterval对比：
        //1 如果比newInterval小：直接加入新结果集合
        //2 如果有overlap，动态改变newInterval为新的区间，继续合并
        //3 如果比newInterval大：加入newInterval到新集合，然后把newInterval更新为当前对象
        //解题思路：画图，先排除两种不重叠的情况，对重叠情况最后处理，
        //需要注意中间值temp的变化，以及最后需要添加最后一个temp
        List<Interval> res=new ArrayList<Interval>();
        Interval temp=newInterval;
        for(int i=0;i<intervals.size();i++){
            
            Interval cur=intervals.get(i);
            if(cur.start>temp.end){
                res.add(temp);
                temp=cur;
            }else{
                if(temp.start>cur.end){
                    res.add(cur);
                }else{
                    int start=Math.min(cur.start,temp.start);
                    int end=Math.max(cur.end,temp.end);
                    Interval newInt=new Interval(start,end);
                    temp=newInt;
                }
            }
            
            
        }
        res.add(temp);
        return res;
    }
}