Maximum sum of the range non-overlapping intervals

问题:

Someone asked me this question:
You are given a list of intervals. You have to design an algorithm to find the sequence of non-overlapping intervals so that the sum of the range of intervals is maximum.

For Example:
If given intervals are:

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

Range is maximized when three intervals

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

are chosen.

Therefore, the answer is 420 (minutes).

回答1:

This is a standard interval scheduling problem.
It can be solved by using dynamic programming.

Algorithm
Let there be n intervals. sum[i] stores maximum sum of interval up to interval i in sorted interval array. The algorithm is as follows

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])

The iteration goes for n steps and in each step, j can be found using binary search, i.e. in log n time. Hence algorithm takes O(n log n) time.

回答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;
        }

    }


}

Heres the working code. Basically this runs in O(n^2) because of the two for loops. But as Shashwat said there are ways to make it run in O(n lg n)