Question

假设我有3支球队，每支球队需要互相比赛一次。如何有效地迭代它们来安排匹配而不重复？

    $teams = range('a', 'c');
    $games = array();
    foreach ($teams as $t1)
        foreach ($teams as $t2)
            if ($t1 != $t2)
                $games[] = $t1 . ' vs. ' . $t2;
    print_r($games);

结果：

    Array
    (
        [0] => a vs. b
        [1] => a vs. c
        [2] => b vs. a  <-- Duplicate of [0]
        [3] => b vs. c
        [4] => c vs. a  <-- Duplicate of [1]
        [5] => c vs. b  <-- Duplicate of [3]
    )

如何有效避免重复匹配？

答案应该产生：

    Array
    (
        [0] => a vs. b
        [1] => a vs. c
        [2] => b vs. c
    )

Answer 1

一般地：

Container C;
for (i = 0; i < C.size; ++i)
{
  for (j = i + 1; j < C.size; ++j)
  {
    /* have unique unordered set {i, j} */
  }
}

如果您想要对角线条目(i, i)，请从j = i开始。

Answer 2

这个解决方案并不像@Kerrek那样优雅，但它的工作方式相同：

$teams = range('a', 'c');
$games = Array();
$matched = Array();

foreach ( $teams as $t1 )
    foreach ( $teams as $t2 )
        if ( $t1 != $t2 )
            if ( !in_array( Array( $t1, $t2 ), $matched ) && !in_array( Array( $t2, $t1 ), $matched ) )
            {

                $games[] = $t1 . ' vs. ' . $t2;
                $matched[] = Array( $t1, $t2 );

            }

print_r( $games );

如何在不重复的情况下有效地迭代数组中的对？

2 个答案: