将null作为整数类型的默认值

Question

An algorithm from Leetcode Discussion uses Binary Search Tree to hold a range of values from an input array to check if this array contains values that are different at most by t, and their indexes are at most k far from each other.

JAVA :

public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
    if (nums == null || nums.length == 0 || k <= 0) {
        return false;
    }

    final TreeSet<Integer> values = new TreeSet<>();
    for (int ind = 0; ind < nums.length; ind++) {

        final Integer floor = values.floor(nums[ind] + t);
        final Integer ceil = values.ceiling(nums[ind] - t);

        if ((floor != null && floor >= nums[ind])
                || (ceil != null && ceil <= nums[ind])) {
            return true;
        }

        values.add(nums[ind]);
        if (ind >= k) {
            values.remove(nums[ind - k]);
        }
    }
    return false;
}

I am struggling to make it work in C#. This code doesn't work as long as LastOrDefault and FirstOrDefault methods return 0 as a default. How to work around to get null as a default?

C# :

public bool ContainsNearbyAlmostDuplicate(int[] nums, int k, int t)
{
    if (nums == null || nums.Length < 2 || k < 1) return false;

    SortedSet<long> set = new SortedSet<long>();

    for (int i = 0; i < nums.Length; i++)
    {
        long l = (long)nums[i];

        long floor = set.LastOrDefault(n => n <= l);
        long ceil = set.FirstOrDefault(n => n >= l);

        if ((l - floor <= t) || (ceil - l <= t))
            return true;

        set.Add(l);
        if (i >= k) set.Remove((long)nums[i - k]);
    }
    return false;
}

Answer 1

One way would be to declare your set with nullable longs

var set = new SortedSet<long?>();

Or you could just not use the FirstOrDefault and do something like:

var greaterOrEqualToOne = set.Where(n => n >= 1);

long? ceil = greaterOrEqualToOne.Any() ? greaterOrEqualToOne.First() : null;

Yet another way, cast them to long? first:

long? ceil = set.Select(n => (long?)n).FirstOrDefault(n => n >= 1);

Answer 2

使用nullable long在LastOrDefault上获取null

SortedSet<long?> set = new SortedSet<long?>();

Answer 3

If you want to retrieve int/long/... values including null, then you need to user Nullable Types

And in your source code will be:

 SortedSet<long?> set = new SortedSet<long?>();

Answer 4

Other answers are technically correct (use nullable types on your set)

SortedSet<long?> set = new SortedSet<long?>();

However, in order to ever receive null value from FirstOrDefault / LastOrDefault you need to pass the int?[] nums to your method, because you fill your set by casting

long l = (long)nums[i];

which won't get you null value from int array even when casted to nullable long? .

In other words - your method needs to get nullable array in order to work with null values.

And if you won't ever be working with nulls on provided array why not just iterate from i = 1 ?

Answer 5

You can cast int/long it to nullable type (?) as other users have stated and then use DefaultIfEmpty to create a default collection that contains null if the collection turns out to be empty.

SortedSet<long?> set = new SortedSet<long?>();

long? floor = set.Where(n => n <= l).DefaultIfEmpty().LastOrDefault();
long? ceil = set.Where(n => n >= l).DefaultIfEmpty().FirstOrDefault();

Answer 6

If you want to declare int,long... values as NULL, you need to declare like this:

int? empId;
if(empId is null)
{
    // do this
}
else
{
    //do this
}

In the same way you have to use:

SortedSet<long?> sortedSet = new SortedSet<long?>();