[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;
}
}