leetcode数组专项习题：消除间隔

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17、消除间隔
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].
题设要求：给定区间的集合，合并所有重叠的区间。例如，给定区间集合[1,3],[2,6],[8,10],[15,18], 返回[1,6],[8,10],[15,18]。
分析：区间本身是左小右大，在合并时，可以借助数轴来思考。
代码：

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;

public class Solution{
    public static void main(String[] args) 
    {
    	Solution sl = new Solution();
    	ArrayList<Interval> intervals= new ArrayList<Interval>(); 
    	Interval i1=new Interval(1,3);
    	Interval i2=new Interval(2,6);
    	Interval i3=new Interval(8,10);
    	Interval i4=new Interval(15,18);   	
    	intervals.add(i1);
    	intervals.add(i2);
    	intervals.add(i3);
    	intervals.add(i4);
    	ArrayList<Interval> result=(ArrayList<Interval>) sl.merge(intervals);
     }

	public ArrayList<Interval> merge(ArrayList<Interval> intervals) 
        {
            ArrayList<Interval> res = new ArrayList<>();
            if(intervals.size() == 0)
                return res;
            
            Collections.sort(intervals, new Comparator<Interval>(){
                public int compare(Interval a, Interval b)
                {
                    return a.start - b.start;
                }
            });;
           
            Interval temp = intervals.get(0);

            if(intervals.size() == 1)
            {
                res.add(temp);
                return res;
            }
            
            for(int i=1; i<intervals.size(); i++)
            {
                if(temp.end >= intervals.get(i).start)
                {
                    temp.end = Math.max(temp.end, intervals.get(i).end);
                }
                else
                {
                    res.add(temp);
                    temp = intervals.get(i);
                }
            }
            res.add(temp);
            return res;
        }
}