间隔列表中非重叠区间范围的最大总和

Question

有人问我这个问题:
给你一个间隔清单.您必须设计一种算法来查找非重叠区间的序列,以便区间范围的总和最大.

例如:
如果给定的间隔是:

["06:00","08:30"],
["09:00","11:00"],
["08:00","09:00"],
["09:00","11:30"],
["10:30","14:00"],
["12:00","14:00"]

三个间隔时范围最大化

[“06:00”, “08:30”],
[“09:00”, “11:30”],
[“12:00”, “14:00”],

被选中.

因此,答案是420(分钟).

Answer 1

这是标准的间隔调度问题.
它可以通过使用动态编程来解决.

算法
让n间隔.在排序的间隔数组中sum[i]存储最大间隔的最大i间隔.算法如下

Sort the intervals in order of their end timings.
sum[0] = 0
For interval i from 1 to n in sorted array
    j = interval in 1 to i-1 whose endtime is less than beginning time of interval i.
    If j exist, then sum[i] = max(sum[j]+duration[i],sum[i-1])
    else sum[i] = max(duration[i],sum[i-1])

迭代用于n步骤,并且在每个步骤中,j可以使用二分搜索找到,即在log n时间上.因此算法需要O(n log n)时间.

Answer 2

public int longestNonOverLappingTI(TimeInterval[] tis){
        Arrays.sort(tis);
        int[] mt = new int[tis.length];
        mt[0] = tis[0].getTime();
        for(int j=1;j<tis.length;j++){
            for(int i=0;i<j;i++){
                int x = tis[j].overlaps(tis[i])?tis[j].getTime():mt[i] + tis[j].getTime();
                mt[j]  = Math.max(x,mt[j]);
            }
        }

        return getMax(mt);
    }


public class TimeInterval implements Comparable <TimeInterval> {
    public int start;
    public int end;
    public TimeInterval(int start,int end){
        this.start = start;
        this.end = end;

    }



    public boolean overlaps(TimeInterval that){
          return !(that.end < this.start || this.end < that.start);
    }

    public int getTime(){
        return end - start;
    }
    @Override
    public int compareTo(TimeInterval timeInterval) {
        if(this.end < timeInterval.end)
            return -1;
        else if( this.end > timeInterval.end)
            return 1;
        else{
            //end timeIntervals are same
            if(this.start < timeInterval.start)
                return -1;
            else if(this.start > timeInterval.start)
                return 1;
            else
                return 0;
        }

    }


}

这是工作代码。基本上，由于有两个 for 循环，它的运行时间为 O(n^2)。但正如 Shashwat 所说，有一些方法可以让它在 O(n lg n) 内运行