我目前正在经历一场脑筋急转弯的老屁，并且被困在动态选择下轮比赛获胜者将晋级的下一场“回合比赛”:
上面的梯子是动态生成的，我想做的是找出下一个比赛ID。我目前已将此作为POC，但如果竞争阶梯要达到64/以上，这是不可持续的:
$ar = [
1 => [
['id' => 1,'name' => 'round1, pair 1'],
['id' => 2,'name' => 'round1, pair 2'],
['id' => 3,'name' => 'round1, pair 3'],
['id' => 4,'name' => 'round1, pair 4'],
],
2 => [
['id' => 5,'name' => 'round2, pair 1'],
['id' => 6,'name' => 'round2, pair 2'],
]
];
$cases = [0, 0, 1, 1, 2, 2];
foreach($ar as $i => $round) {
foreach($round as $_i => $r) {
echo $r['name'] . " & NEXT_MATCH_ID::> " . $ar[($i + 1)][$cases[$_i]]['id'] . "<br /> ";
}
}
是否有一种更简化的方式来实现上述目的，例如，没有硬编码的变量(
$cases)。本质上，“匹配/对”的数量被减半，因为梯子应为:
4-> 2-> 1。上面的代码会生成正确的ID，但是它是不可扩展的或动态的；
round1, pair 1 & NEXT_MATCH_ID::> 5
round1, pair 2 & NEXT_MATCH_ID::> 5
round1, pair 3 & NEXT_MATCH_ID::> 6
round1, pair 4 & NEXT_MATCH_ID::> 6
round2, pair 1 & NEXT_MATCH_ID::> ...
round2, pair 2 & NEXT_MATCH_ID::> ...
//......etc etc...
上述代码的Demo/ Example(如果需要)。
注释
4, 6, 8, 10, 12, 14, 16, 18....32, 34...64...etc。 1比赛)，因为没有进一步的回合可以进行。 (很容易受if($i == count($rounds)) {.... do not continue...限制)。 lastId + 1。 最佳答案
只是数学
请记住，每个回合都包含球队的pow(2, Rounds - Round + 1)和比赛的pow(2, Rounds - Round)。只需将其概括为几何级数即可。
在回合$round之前进行的比赛次数是geometric progression 2^(rounds-1) + 2^(rounds-2) + ... 2^(rounds - round + 1)与a=2^(rounds-1)，r=1/2和n=round-1。它的总和是2^(rounds) - 2^(rounds+1-round)。
因此，匹配ID和下一个匹配ID只是三个参数的函数:pairnum，round和rounds。我将其计算转移到函数getMatchId和getNextId中。
示例
<?php
// just matchesInPreviousRounds + parnum
function getMatchId($pairnum, $round, $rounds) {
// matchesInPreviousRounds - is a sum of a geometric progression
// 2^(rounds-1) + 2^(rounds-2) + ... 2^(rounds - round + 1)
// with a=2^(rounds-1), r=1/2, n = round-1
// its sum is 2^(rounds) - 2^(rounds+1-round)
$inPreviousRounds = $round > 1 ? (pow(2, $rounds) - pow(2, $rounds + 1 - $round)) : 0;
$id = $inPreviousRounds + $pairnum;
return (int)$id;
}
// next id is last id of a round + half a pairnum.
function getNextId($pairnum, $round, $rounds) {
if($round === $rounds) {
return false;
}
$matchesInThisAndPreviousRounds = pow(2, $rounds) - pow(2, $rounds - $round);
$nextid = $matchesInThisAndPreviousRounds + ceil($pairnum / 2);
return (int)$nextid;
}
$divide = 64; // for 1/64 at the start
$power = round(log($divide) / log(2)); // get 6 for 64
$rounds = (int) $power + 1;
for($round = 1; $round <= $rounds; $round++) {
// every round contains 2^($rounds - $round + 1) of teams
// and has 2^($rounds - $round) of matches
$teamsLeft = pow(2, $rounds - $round + 1);
$pairsLeft = pow(2, $rounds - $round);
for($pairnum = 1; $pairnum <= $pairsLeft; $pairnum++) {
$id = getMatchId($pairnum, $round, $rounds);
$nextid = getNextId($pairnum, $round, $rounds);
echo "Round $round, pair $pairnum, id $id ";
echo "winner goes to " . $nextid ? $nextid : "A BAR" . "\n";
}
}
其结果
Round 1, pair 1, id 1, winner goes to 65
Round 1, pair 2, id 2, winner goes to 65
...
Round 1, pair 62, id 62, winner goes to 95
Round 1, pair 63, id 63, winner goes to 96
Round 1, pair 64, id 64, winner goes to 96
Round 2, pair 1, id 65, winner goes to 97
Round 2, pair 2, id 66, winner goes to 97
...
Round 2, pair 29, id 93, winner goes to 111
Round 2, pair 30, id 94, winner goes to 111
Round 2, pair 31, id 95, winner goes to 112
Round 2, pair 32, id 96, winner goes to 112
Round 3, pair 1, id 97, winner goes to 113
Round 3, pair 2, id 98, winner goes to 113
...
Round 3, pair 13, id 109, winner goes to 119
Round 3, pair 14, id 110, winner goes to 119
Round 3, pair 15, id 111, winner goes to 120
Round 3, pair 16, id 112, winner goes to 120
Round 4, pair 1, id 113, winner goes to 121
Round 4, pair 2, id 114, winner goes to 121
...
Round 4, pair 7, id 119, winner goes to 124
Round 4, pair 8, id 120, winner goes to 124
Round 5, pair 1, id 121, winner goes to 125
Round 5, pair 2, id 122, winner goes to 125
Round 5, pair 3, id 123, winner goes to 126
Round 5, pair 4, id 124, winner goes to 126
Round 6, pair 1, id 125, winner goes to 127
Round 6, pair 2, id 126, winner goes to 127
Round 7, pair 1, id 127, winner goes to A BAR
关于php - 动态获取/排序下一个多维数组元素，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/42239239/