如何在循环中从int的HashSet列表中删除项目

Question

I have the following lists:

        var set = new List<HashSet<int>>()
        {
            new HashSet<int>() { 1,2,3,4},
            new HashSet<int>() { 1,2,3,5},
            new HashSet<int>() { 1,2,4,5},
            new HashSet<int>() {2,3,4,5}
        };

        var subSet = new List<HashSet<int>>()
        {
            new HashSet<int>() { 1,2,3},
            new HashSet<int>() { 1,2,4},
        };

I want to remove from set the subSet.Count() items, which isProperSubSet, and the result must be:

        var result= new List<HashSet<int>>()
        {
            new HashSet<int>() { 1,2,3,5},             
            new HashSet<int>() {2,3,4,5}
        };

How can I do it?

I tried like this, but I get an index error (must be non negative and less then the collection):

for(int j=set.Count-1;j-->0;)
        {
            for (int i = subSet.Count-1;i-->0;)
            {
                if (subSet[i].IsProperSubsetOf(set[j]))
                {
                    subSet.RemoveAt(i);
                    set.RemoveAt(j);
                }

            }
        }

Answer 1

First, you should not remove elements from subSet or the other elements of set which the subset is not a proper subSet won't be removed.

Second, juharr said, you have to add a break inside the if :

for(int j=set.Count-1;j-->0;)
    {
        for (int i = subSet.Count-1;i-->0;)
        {
            if (subSet[i].IsProperSubsetOf(set[j]))
            {
                set.RemoveAt(j);
                break;
            }

        }
    }

Answer 2

Basically you need to break out of the inner loop once you remove a value from set .

for(int j=set.Count-1; j >= 0; j--)
{
    for (int i = subSet.Count-1; i >= 0; i--)
    {
        if (subSet[i].IsProperSubsetOf(set[j]))
        {
            subSet.RemoveAt(i);
            set.RemoveAt(j);
            break;
        }
     }
}

That's because you might still be iterating the inner loop and now your j will actually reference the wrong position (which might be beyond the new length of the list).