Question

我想编写一个程序，以二进制等值的十进制数找到两个1之间的最大间隙。例如，对于100101：间隙为2，对于10101：间隙为1.

<?php
$numberGiven = 251;
$binaryForm = decbin($numberGiven);

$status = false;
$count = 0;
for($i = 0; $i < strlen($binaryForm); $i++)
{
    var_dump($binaryForm[$i]);
    if($binaryForm[$i] == 1)
    {
        $status = false;
        $count = 0;
    }
    else
    {
        $status = true;
        $count += 1;
    }
}
echo "count = " . $count . "<br>";
echo $binaryForm;
?>

但我没有成功..

Answer 1

您目前正在做的是每次找到1时重置计数。

您需要跟踪当前的最大值：

$count = 0;
$maxCount = 0;

以及您设置$count = 0的位置，您还应该

if ($count > $maxCount)
  $maxCount = $count;
$count = 0;

然后

echo "count = " . $maxCount . "<br>";

总之：

<?php
$numberGiven = 251;
$binaryForm = decbin($numberGiven);

$status = false;    
$count = 0;
$maxCount = 0;

for($i = 0; $i < strlen($binaryForm); $i++)
{
  // Don't count leading zeroes.
  if ($status == false && $binaryForm[$i] == 0) continue;
  $status = true;

  var_dump($binaryForm[$i]);

  // We've found a 1. Remember the count.
  if($binaryForm[$i] == 1)
  {
    if ($count > $maxCount)
      $maxCount = $count;
    $count = 0;
  }
  // We found a 0. Add one to count.
  else
  {
    $count += 1;
  }
}

echo "count = " . $count . "<br>";
echo $binaryForm;
?>

免责声明： 未经过测试的代码

Answer 2

我会使用binary right-shift operator >>并迭代地移位1位，并检查当前最右边的位是1，直到我检查了所有位。如果找到1，则会计算之前1之间的差距：

foreach(array(5,17,25,1223243) as $number) {
    $lastpos = -1; 
    $gap = -1; // means there are zero or excatly one '1's

    // PHP_INT_SIZE contains the number of bytes an integer
    // will consume on your system. The value * 8 is the number of bits.
    for($pos=0; $pos < PHP_INT_SIZE * 8; $pos++) {
        if(($number >> $pos) & 1) {
            if($lastpos !== -1) {
                $gap = max($gap, $pos - $lastpos -1);
            }
            $lastpos = $pos;
        }
    }   

    echo "$number " . decbin($number) . "  ";
    echo "max gap: {$gap}\n";
}

输出：

5 101  max gap: 1
17 10001  max gap: 3
25 11001  max gap: 2
1223243 100101010101001001011  max gap: 2

Answer 3

使用正则表达式查找＆＃34; 0＆＃34;组，按长度降序排序，取第一个，得到长度：

$numberGiven = 251;
$binaryForm = decbin($numberGiven);

preg_match_all('/(0+)/', $binaryForm, $matches);
$matches = $matches[0];

rsort($matches, SORT_STRING);
echo 'count = ' . strlen($matches[0]) . '<br>';
echo $binaryForm;

更新：基于Mark Baker的评论。

更新＃2：由afeijo在下面的评论中提出，上述内容并不排除结尾零。以下是解决方案：

preg_match_all('/(0+)1/', $binaryForm, $matches);
$matches = $matches[1];

两位1之间的最大间隙，二进制当量十进制数

3 个答案: