在数组中找到所有可能的两个组合集合

Question

Given an array of size n which will be even always, I need to output all possible combinations of 2 pair elements.

I have this:

let teams = ["team1","team2","team3","team4"]

what I tried:

function pairs(arr) {
    var res = [],
        l = arr.length;
    for(var i=0; i<l; ++i)
        for(var j=i+1; j<l; ++j)
            res.push([arr[i], arr[j]]);
    return res;
}

pairs(teams).forEach(function(pair){
    console.log(pair);
});

//outputs

[ 'team1', 'team2' ]
[ 'team1', 'team3' ]
[ 'team1', 'team4' ]
[ 'team2', 'team3' ]
[ 'team2', 'team4' ]
[ 'team3', 'team4' ]

This seems fine but some are missing for example I don't see the following :

[ 'team2', 'team1' ] 
[ 'team3', 'team1' ]
[ 'team3', 'team2' ]
[ 'team4', 'team1' ]
[ 'team4', 'team2' ]
[ 'team4', 'team3' ]

So my expected output is the above and previous result in the code I posted above.

Notice that I don't need them in particular order as long as I get all the possible combinations.

Answer 1

Functional JS:

 let teams = ["team1", "team2", "team3", "team4"]; var result = teams.flatMap( (v) => teams. filter( w => v!=w ). map ( w => [v, w] ) ); console.log(result);

Answer 2

If you intend to do more combinatorics than this simple example, then you can use the js-combinatorics package. For example:

const Combinatorics = require('js-combinatorics');

const teams = ['team1', 'team2', 'team3', 'team4'];

const matches = Combinatorics.permutation(teams, 2).toArray();

console.log(matches);

Output is:

[
  [ 'team1', 'team2' ],
  [ 'team2', 'team1' ],
  [ 'team1', 'team3' ],
  [ 'team3', 'team1' ],
  [ 'team2', 'team3' ],
  [ 'team3', 'team2' ],
  [ 'team1', 'team4' ],
  [ 'team4', 'team1' ],
  [ 'team2', 'team4' ],
  [ 'team4', 'team2' ],
  [ 'team3', 'team4' ],
  [ 'team4', 'team3' ]
]

Answer 3

If you want to get all the possible combinations including reverse duplicates, you need to iterate j loop from 0 to array's length as well.

for(let i = 0; i < l; i++) 
  for(let j = 0; j < l; j++) {
    if(i == j) continue; // skip the same index
    res.push([arr[i], arr[j]]); 
  }
}

Answer 4

for(var i=0; i<l; ++i)
    {
        for(var j=0; j<l; ++j)
        {
            if(i!=j)
            res.push([arr[i], arr[j]]);
        }
    }

Above code will generate all pairs

Answer 5

array.flatMap(x => array.map(y => x !== y ? x + ' ' + y : null)).filter(x => x)