What's the quickest way to find all possible pairs in list?

问题

Basically I have list of players, and I want to pair them up so that each player will play everyone once. What's the quickest way to find this data?

回答1:

assuming that players do not appear in the list twice, a double for loop is very quick:

for (int i=0, i <= playerList.Count - 2, i++)
    for (int j=i+1, j <= playerList.Count - 1, j++)
        //add a new pairing of player i and j

回答2:

Such tournament schedule is often called round-robin. In wikipedia, there's also an example of possible scheduling algorithm.

回答3:

I put together 2 implementations to compare performance with. The very naive version 1 is about 50% slower than the version 2. That's not to say that nothing faster exists.

class Program
{
    class Player
    {
        public string Name { get; set; }

        public Player(string name)
        {
            Name = name;
        }
    }

    class Match
    {
        public readonly Player Player1;
        public readonly Player Player2;

        public Match(Player player1, Player player2)
        {
            Player1 = player1;
            Player2 = player2;
        }

        public override string ToString()
        {
            return string.Format("{0} vs. {1}", Player1.Name, Player2.Name);
        }
    }

    static readonly List<Player> _players = new List<Player>()
    {
        new Player("John"),
        new Player("Lisa"),
        new Player("Matt"),
        new Player("Dan"),
        new Player("Steve"),
        new Player("Sarah"),
        new Player("Tim")
    };

    static void Main(string[] args)
    {
        const int count = 1000000;

        {
            var v1 = V1();
            var sw = Stopwatch.StartNew();
            for (int i = 0; i < count; i++)
            {
                v1 = V1();
            }
            Console.WriteLine(v1);
            Console.WriteLine(sw.Elapsed);
        }

        {
            var v2 = V2();
            var sw = Stopwatch.StartNew();
            for (int i = 0; i < count; i++)
            {
                v2 = V2();
            }
            Console.WriteLine(v2);
            Console.WriteLine(sw.Elapsed);
        }


        Console.ReadLine();
    }

    static List<Match> V1()
    {
        var challengers = new List<Player>(_players);
        var matches = new List<Match>();
        foreach (var player in _players)
        {
            challengers.Remove(player);
            foreach (var challenger in challengers)
            {
                matches.Add(new Match(player, challenger));
            }
        }
        return matches;
    }

    static List<Match> V2()
    {
        var matches = new List<Match>();
        for (int i = 0; i < _players.Count; i++)
        {
            for (int j = i + 1; j < _players.Count; j++)
            {
                matches.Add(new Match(_players[i], _players[j]));
            }
        }
        return matches;
    }
}

回答4:

A simple divide and conquer algorithm:

If there are only two people: Pair them and return them.
Otherwise:
1. Split the group in two groups of equal size.
2. Find all pairings in each group using this algorithm recursively.
3. Join the two lists.
  
  E.g. [[(a,b)]] and [[(c,d)]] becomes [[(a,b),(c,d)]].
4. Find pairs across of the two groups, by rotating group two.
  
  E.g. [[(a,c),(b,d)],[(a,d),(b,c)]]
5. Return (3) + (4)

This algorithm runs in O(n^2) time, which is optimal, as it generates (n-1) rounds of n/2 pairings.

For eight players you would get 7 rounds:

[(a,b), (c,d), (e,f), (g,h)]
[(a,c), (b,d), (e,g), (f,h)]
[(a,d), (b,c), (e,h), (f,g)]
[(a,e), (b,f), (c,g), (e,h)]
[(a,f), (b,g), (c,h), (e,e)]
[(a,g), (b,h), (c,e), (e,f)]
[(a,h), (b,e), (c,f), (e,g)]

来源：https://stackoverflow.com/questions/5045137/whats-the-quickest-way-to-find-all-possible-pairs-in-list

标签

algorithm

language-agnostic

round-robin