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].

复用 Insert Interval代码即可。

/**
 * 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 ArrayList<Interval> merge(ArrayList<Interval> intervals) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ArrayList<Interval> results = new ArrayList<Interval>();
        if(intervals == null || intervals.size() == 0) return intervals;
        for(int i = 0; i < intervals.size(); i ++)
        {
            results = insert(results,intervals.get(i));
        }
        return results;
    }

    public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ArrayList<Interval> results = new ArrayList<Interval>();
        boolean sig = false;
        for(int i = 0; i < intervals.size(); i ++)
        {
            if(intervals.get(i).start > newInterval.end)
            {
                if(!sig)
                {
                    results.add(newInterval);
                    sig = true;
                }
                results.add(intervals.get(i));
            }
            else
            {
                newInterval.end = Math.max(newInterval.end, intervals.get(i).end);
            }
            if(intervals.get(i).end < newInterval.start)
            {
                results.add(intervals.get(i));    
            }
            else
            {
                newInterval.start = Math.min(newInterval.start, intervals.get(i).start);
            }
        }
         if(!sig)
        {
            results.add(newInterval);
        }
        return results;
    }
}

 第二遍：

public class Solution {
    public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ArrayList<Interval> result = new ArrayList<Interval>();
        if(intervals == null || intervals.size() == 0)return result;
        PriorityQueue<Interval> pq = new PriorityQueue<Interval>(intervals.size(), new Comparator<Interval>(){
            public int compare(Interval a, Interval b){
                return a.start > b.start ? 1 : (a.start == b.start ? 0 : -1);
            }
        });
        for(Interval interval:intervals){
              pq.add(interval);
        }
        result.add(pq.poll());//poll out the minimum element in the queue
        int max_end = result.get(0).end;
        while(pq.size() != 0){
            Interval tmp = pq.poll();
            if(tmp.start <= max_end){
                max_end = max_end > tmp.end ? max_end : tmp.end;
                result.get(result.size() - 1).end = max_end;
            }else{
                result.add(tmp);
                max_end = tmp.end;
            }
        }
        return result;
    }
}

先把interval 按start time 排序，再添加。 保存当前的最大end time。