對於最少8支球隊和最多18支球隊的比賽,我必須確定比賽日曆。錦標賽有17輪或比賽日。所以每個團隊每個比賽日必須遇到另一個團隊。如果只有不到18支球隊的遭遇可以重複,那麼一支球隊可以不止一次對陣另一個球隊。我應該如何用Javascript解決這個組合場景?
This is an example for 18 teams tournament.And this would be a case for less than 18 teams fixture, here in particular 9 teams。
所以,我得做排列,然後安排他們在不同的回合。我已經試過:
組合:
function k_combinations(set, k) {
var i, j, combs, head, tailcombs;
if (k > set.length || k <= 0) {
return [];
}
if (k == set.length) {
return [set];
}
if (k == 1) {
combs = [];
for (i = 0; i < set.length; i++) {
combs.push([set[i]]);
}
return combs;
}
combs = [];
for (i = 0; i < set.length - k + 1; i++) {
head = set.slice(i, i+1);
tailcombs = k_combinations(set.slice(i + 1), k - 1);
for (j = 0; j < tailcombs.length; j++) {
combs.push(head.concat(tailcombs[j]));
}
}
return combs;
}
var teams = [ {name: 'Real Madrid'},
{name: 'Las Palmas'},
{name: 'Alavés'},
{name: 'Valencia'},
{name: 'Sevilla'},
{name: 'Betis'},
{name: 'Córdoba'},
{name: 'Deportivo'},
{name: 'Atlético de Madrid'},
{name: 'Levante'},
{name: 'Rayo Vallecano'},
{name: 'Athletic Bilbao'},
{name: 'Osasuna'},
{name: 'Zaragoza'},
{name: 'Villareal'},
{name: 'Racing de Santander'},
{name: 'Espanyol'},
{name: 'Cádiz'},
];
// Compute whole encounters combinations.
var seasonMatches = k_combinations(teams,2);
大紅大紫的組合安排:
var calendar = {};
for (var i = 0; i<17; i++) {
calendar[i+1] = [];
}
var encounters = seasonMatches;
for (var i = 0; i<Object.keys(calendar).length; i++) {
encounters.map(function (match,index) {
if (! _.any(calendar, function (m) {
return m[0].name === match[0].name || m[1].name === match[1].name || m[0].name === match[1].name || m[1].name === match[0].name;
})) {
calendar[i+1].push(match);
}
});
}
我使用lodash簡化任何遭遇是否存在等檢查在前面回合。
我得到的問題是,這種方式我得到了相同的日曆中的每一個相遇。而且,如果我在seasonMatch中添加拼接,則每輪最終都會有不同的匹配。
I've got a fiddle with this example shown above. 我應該如何解決這個問題?
感謝的人!晚會很晚,但我很感激 – diegoaguilar