题目
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
这道题目比较简单，就是逻辑结构有点繁琐。
代码
/**
* 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) {
if(intervals.size()<=0)
{
intervals.add(newInterval);
return intervals;
}
for(int i=0;i<intervals.size();i++)
{
int astart = intervals.get(i).start;
int aend = intervals.get(i).end;
int bstart = newInterval.start;
int bend = newInterval.end;
if(bstart<astart)
{
if(bend<astart)
{
intervals.add(i,newInterval);
return intervals;
}
else{
if(bend<=aend)
{
intervals.get(i).start = bstart;
return intervals;
}
else
{
intervals.remove(i);
i--;
System.out.println(i);
continue;
}
}
}
else{
if(bstart<=aend){
if(bend<=aend)
{
return intervals;
}
else
{
newInterval.start = astart;
intervals.remove(i);
i--;
continue;
}
}
else{
continue;
}
}
}
intervals.add(newInterval);
return intervals;
}
}