C# 找到给定列表中的最小对

让我们有一个int数组[{0,1}，{1,2}，{4,5}]的列表，他们是成对的朋友 我们怎样才能找到其他对的朋友呢。好友列表将是0,1,2和4,5，并用c打印。 如果该数字存在于int数组对中，那么它将被添加到中，并用C打印 例如；列表有[{0,1}，{1,2}，{4,5}] 然后可能的打印将为0,1,2和4,5

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让0,1是朋友，1,2是朋友，4,5是朋友，那么0,1,2是朋友，4,5是朋友

这能满足您的需要吗

List<int[]> pairsOfFriends = new List<int[]>
{
    new int[] {0, 1},
    new int[] {1, 2},
    new int[] {4, 5},
};
Dictionary<int, List<int>> friendsLists = new Dictionary<int, List<int>>();
pairsOfFriends.ForEach(pairOfFriends =>
{
    int friendA = pairOfFriends[0];
    int friendB = pairOfFriends[1];
    //if friendA has a friends list, then friendA wants to share his friends list with friendB!
    if (friendsLists.ContainsKey(friendA))
    {
        var friendsListA = friendsLists[friendA];
        //if friendB has a friend list, they also want to share all of their friends with friendA.
        if (friendsLists.ContainsKey(friendB))
        {
            //friendA copies all of friendB's friends into his own friends list
            friendsListA.AddRange(friendsLists[friendB]);
            //friendB and friendA then share friendA's friends list so that they share the same friends!
            friendsLists[friendB].ForEach(friendBsFriend =>
            {
                friendsLists.Remove(friendBsFriend);
                friendsLists.Add(friendBsFriend,friendsListA);
            });
        }
        else
        {
            //if friendB doesn't have any friends, then friendA shares all of his friends with friendB and then adds friendB to his own friends list
            friendsLists.Add(friendB, friendsListA);
            friendsListA.Add(friendB);
        }
        //if friendB has a friends list, and friendA doesnt then friendB adds friendA to his friends list and then shares his friends list with friendA.
    }
    else if (friendsLists.ContainsKey(friendB))
    {
        var friendsListB = friendsLists[friendB];
        friendsLists.Add(friendA, friendsListB);
        friendsListB.Add(friendA);
    }
    //if neither friendB or friendA have a friends list then friend a makes a new friends list, adds himself and friendB to it and then shares the friends list with friend B
    else
    {
        friendsLists.Add(friendA, new List<int> {friendA, friendB});
        friendsLists.Add(friendB, friendsLists[friendA]);
    }
});
friendsLists.Values.Distinct().ToList().ForEach(f =>
{
    Console.Write("("+string.Join(", ", f)+") ");
});

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如果您向我们展示您已经尝试过的内容，您会得到更好的响应。发布的值没有任何意义。我在函数中传递了数组列表，如：new int[3,2]{{0,1}，{1,2}，{3,4}；试着找到配对这是完全相同的例子…啊，我想这个问题是关于如何找到循环。这里{0,1}和{1,2}组合形成{0,1,2}。