对于最少8支球队和最多18支球队的比赛,我必须确定比赛日历。锦标赛有17轮或比赛日。所以每个团队每个比赛日必须遇到另一个团队。如果只有不到18支球队的遭遇可以重复,那么一支球队可以不止一次对阵另一个球队。我应该如何用Javascript解决这个组合场景?
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,然后安排他们在不同的回合。我已经试过:
组合:
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);
大红大紫的组合安排:
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简化任何遭遇是否存在等检查在前面回合。
我得到的问题是,这种方式我得到了相同的日历中的每一个相遇。而且,如果我在seasonMatch中添加拼接,则每轮最终都会有不同的匹配。
I've got a fiddle with this example shown above. 我应该如何解决这个问题?
感谢的人!晚会很晚,但我很感激 – diegoaguilar