【LeetCode】-Merge Intervals

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]

.

/**
* 解题思路：
* 1.将所有区间按左端点的大小升序排列；
* 2.遍历区间列表，比较当前区间与前一个区间的关系：
* a.无交集，直接添加到输出列表;
* b.有交集：如果是包含关系，不作处理，否则更新前一个区间的右端点。
*
* 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 class MyComparator implements Comparator<Interval>{

@Override
public int compare(Interval arg0, Interval arg1) {
// TODO Auto-generated method stub
if( arg0.start==arg1.start )
return 0;
return arg0.start<arg1.start? -1:1;
}

}

public List<Interval> merge(List<Interval> intervals) {
if( intervals==null )
return null;
Collections.sort( intervals, new MyComparator());
List<Interval> list = new ArrayList<Interval>();
Iterator<Interval> itr = intervals.iterator();
while( itr.hasNext() ){
Interval interval1 = itr.next();
if(list.isEmpty()){
list.add( interval1 );
}else{
Interval interval2 = list.get( list.size()-1 );
if( interval2.end<interval1.start ){
list.add( interval1 );
}else if( interval2.end<=interval1.end ){
interval2.end = interval1.end;
}
}
}
return list;
}
}