间隔问题，合并间隔(merge interval)，插入间隔(insert interval)

Merge Interval:

Given a collection of intervals, merge all overlapping intervals.

For example,
Given

[1,3],[2,6],[8,10],[15,18]

,
return

[1,6],[8,10],[15,18]

.

算法分析：首先要对给定序列排序。然后再去遍历合并。

class Interval
{
int start;
int end;
Interval()
{
start = 0;
end = 0;
}
Interval(int s, int e)
{
start = s;
end = e;
}
}
public class MergeIntervals
{
public List<Interval> merge(List<Interval> intervals)
{
List<Interval> res = new ArrayList<>();
if(intervals == null || intervals.size() == 0)
{
return res;
}
Collections.sort(intervals, new Comparator<Interval>()//自定义比较方法，外部实现Comparator接口
{
public int compare(Interval i1, Interval i2)
{
if(i1.start != i2.start)
{
return i1.start - i2.start;
}
else
{
return i1.end - i2.end;
}
}
});
Interval pre = intervals.get(0);
for(int i = 0; i < intervals.size(); i ++)
{
Interval curr = intervals.get(i);
if(curr.start > pre.end)
{
res.add(pre);
pre = curr;
}
else
{
Interval merged = new Interval(pre.start, Math.max(pre.end, curr.end));
pre = merged;
}
}
res.add(pre);
return res;
}
}

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]

.

给定间隔序列和插入间隔，默认给定间隔序列有序，返回合并后的不重叠的间隔序列。是上一道题的变形。

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