Question:

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].

---------------------------------------

Solution:

public class Solution {
class IntervalAsc implements Comparator<Interval>
{
public int compare (Interval o1, Interval o2)
{
if (o1.start != o2.start)
return o1.start < o2.start ? -1 : 1;
else if (o1.end != o2.end)
return o1.end < o2.end ? -1 : 1;
else
return 0;
}} public ArrayList<Interval> merge (ArrayList<Interval> intervals)
{
Collections.sort(intervals, new IntervalAsc());
ArrayList<Interval> ret = new ArrayList<Interval>();
int n = intervals.size();
if(n==0) return ret;
Interval last=intervals.get(0);
for(int i=1;i<n;i++){
if(intervals.get(i).start>last.end){
ret.add(new Interval(last.start,last.end));
last=intervals.get(i);
}else{
last.end=Math.max(intervals.get(i).end, last.end);
}
}
ret.add(last);
return ret;
} }

需要注意的以下几点:

  1. 这里需要new一个Interval,再放进ArrayList result里。刚开始我就是直接把last放进了result里,忘记了last只是一个类似于指针一样的东西。
  2. 第一次使用到了内部类。更多的内部类的知识看下一篇博客。
  3. Collections.sort的用法。(compare函数的重写)

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