如何使程序帐户中的数组列表中不存在的索引

Question

My program is supposed to be passed 2 array-lists, arrivals and duration, and should return the number of events that can basically take place without overlap. However, a-lot of testcases are not being passed because the program does not count the duration after the final arrival of the arraylist.

Below is what I have so far:

class Results {
    public static int maxEvents(List<Integer> arrival, List<Integer> duration) {
        int counter = 0;
        if (arrival.size() == 0) {
            counter = 0;
        } else {
            for (int i = 0; i < arrival.size() - 1; i++) {
                if (arrival.get(i) + duration.get(i) <= arrival.get(i + 1)) {
                    counter++;
                } else if (arrival.get(i) == arrival.get(i + 1)) {
                    counter++;
                    i++;
                }
            }
        }
        return counter;
    }
}

I have tried to the following but not only did it not pass the current failed testcases but also caused alot of the ones that were successful to fail as well:

else if (i == arrival.size()-2) {
    counter++;
}

To show the expected testcase is the following:

arrivals = [1,3,3,5,7]
duration = [2,2,1,2,1]

The first person arrives at 1, presents for 2 hours, then leaves. 2 people arrive at 3 but only one is allowed to present for 2 or 1 hours. The next person arrives at 5, presents for 2 hours. The final person arrives at 7, and presents for 1 hour. The answer output should be 4 as 4 people are able to present.

Answer 1

Assuming your input looks something like this:

List<Integer> arrival = new ArrayList<Integer>();
arrival.add(10);
arrival.add(12);
arrival.add(15);
arrival.add(17);
arrival.add(20);

List<Integer> duration = new ArrayList<Integer>();
duration.add(3);
duration.add(1);
duration.add(2);
duration.add(4);
duration.add(3);

I would recommend adding a value which stores the closest available time so you don't need to access two elements of your list at each step:

public static int maxEvents(List<Integer> arrival, List<Integer> duration)
{
    int counter = 0;
    Integer nextTime = 0;

    for (int i = 0; i < arrival.size(); i++)
    {
        if(nextTime <= arrival.get(i))
        {
            counter++;
            nextTime = arrival.get(i) + duration.get(i);
        }
    }

    return counter;
}

This would make sure that each element in your list gets properly tested.

Answer 2

The rules seem to be:

• An arrival must end at most at the next arrival.
• Or the next arrival must be the same, then the next arrival is skipped (not considering its duration).

Expected: - The data is sorted, by arrival.

These rules are slighty weird. For instance your algorithm does not take into account when an arrival is repeated more than two times - but that is your homework.

What is easiest to handle the last case after the loop.

      int i;
      for (i = 0; i < arrival.size()-1; i++) {
          if (arrival.get(i) + duration.get(i) <= arrival.get(i+1)) {
              counter++;
          } else if (arrival.get(i).intValue() == arrival.get(i+1)) {
              counter++;
              i++;
          } 
      }
      if (i < arrival.size()) {
         counter++;
      }

When get(i) returns an Integer <= will ensure the int value is taken (is called automatic unboxing and uses Integer.intValue() . For == simply the Integer are compared, and in general only the integers -128 .. 127 will be cached as unique object. Probably Integer.valueOf(200) != Integer.valueOf(200) .

To prevent such a pitfall, use plural names for the lists: arrivals and durations , and use help variables:

int arrival = arrivals.get(i);
int nextArrival = arrivals.get(i+1);