MYSQL中的序列
MYSQL中的序列
提问于 2016-06-07 06:56:55
在MYSQL中有没有办法跟踪连续数字的子序列？
我想计算数字序列中最高的连续子序列：
1,2,3,4,6,7 => 4 (1-4)
1,2,3,4,5,6,8 => 6 (1-6)
1,2,3,5 => 3 (1-3)
1,2,3,5,6,7,8 => 4 (5-8)
1,2,4,5,6,8,9 => 3 (4-6)回答 2
Stack Overflow用户
发布于 2016-06-07 09:08:01
根据你的问题，这段代码应该可以工作：
function findHighestSeriesCount( $seriesArray )
{
if ( 1 >= count( $seriesArray ) )
{
# catch the special case when array has only 0 or 1 element
return count( $seriesArray );
}
$highestCount = -1;
$currentCount = 1;
$lastElement = $seriesArray [0];
$getCorrectHighValue = function() use ( &$highestCount, &$currentCount )
{
return $highestCount < $currentCount ? $currentCount : $highestCount;
};
for ( $arrayIndex = 1 ; $arrayIndex < count( $seriesArray ) ; $arrayIndex++ )
{
if ( $lastElement + 1 == $seriesArray [$arrayIndex] )
{
$currentCount++;
}
else
{
$highestCount = $getCorrectHighValue();
$currentCount = 1;
}
$lastElement = $seriesArray [$arrayIndex];
}
return $getCorrectHighValue();
}测试脚本：
$seriesArray = [1,2,3,4,6,7];
echo sprintf( "array=%s - highest count=%d<br>", implode( ",", $seriesArray ),
findHighestSeriesCount( $seriesArray ) );
$seriesArray = [1,2,3,4,5,6,8];
echo sprintf( "array=%s - highest count=%d<br>", implode( ",", $seriesArray ),
findHighestSeriesCount( $seriesArray ) );
$seriesArray = [1,2,3,5];
echo sprintf( "array=%s - highest count=%d<br>", implode( ",", $seriesArray ),
findHighestSeriesCount( $seriesArray ) );
$seriesArray = [1,2,3,5,6,7,8];
echo sprintf( "array=%s - highest count=%d<br>", implode( ",", $seriesArray ),
findHighestSeriesCount( $seriesArray ) );
$seriesArray = [1,2,4,5,6,8,9];
echo sprintf( "array=%s - highest count=%d<br>", implode( ",", $seriesArray ),
findHighestSeriesCount( $seriesArray ) );输出：
array=1,2,3,4,6,7 - highest count=4
array=1,2,3,4,5,6,8 - highest count=6
array=1,2,3,5 - highest count=3
array=1,2,3,5,6,7,8 - highest count=4
array=1,2,4,5,6,8,9 - highest count=3Stack Overflow用户
发布于 2016-06-07 09:31:58
假设数据存储为一列，您可以通过查找最长的连续序列来获得最大值。这个想法很简单，就是减去另一个序列。。。在这些值是恒定的情况下，你知道你有一个序列。
其余的只是聚合。这看起来像这样：
select count(*)
from (select t.*, (@rn := @rn + 1) as rn
from t cross join (select @rn := 0) params
) t
group by (col - rn)
order by count(*) desc
limit 1;页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/37668054
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