用于排序数组中的天花板的 PHP 程序
给定一个排序数组和一个值 x,x 的上限是数组中大于或等于 x 的最小元素,下限是小于或等于 x 的最大元素。假设数组按非降序排序。编写有效的函数来找到 x 的下限和上限。
例子 :
For example, let the input array be {1, 2, 8, 10, 10, 12, 19}
For x = 0: floor doesn't exist in array, ceil = 1
For x = 1: floor = 1, ceil = 1
For x = 5: floor = 2, ceil = 8
For x = 20: floor = 19, ceil doesn't exist in array
在下面的方法中,我们只实现了天花板搜索功能。楼层搜索可以以相同的方式实现。
方法一(线性搜索)
搜索 x 上限的算法:
1) 如果 x 小于或等于数组中的第一个元素,则返回 0(第一个元素的索引)
2) Else 线性搜索索引 i,使得 x 位于 arr[i] 和 arr[i+1] 之间。
3) 如果我们在第 2 步中没有找到索引 i,则返回 -1
PHP
= $x)
return $i + 1;
}
// If we reach here then x is greater
// than the last element of the array,
// return -1 in this case
return -1;
}
// Driver Code
$arr = array(1, 2, 8, 10, 10, 12, 19);
$n = sizeof($arr);
$x = 3;
$index = ceilSearch($arr, 0, $n - 1, $x);
if($index == -1)
echo("Ceiling of " . $x .
" doesn't exist in array ");
else
echo("ceiling of " . $x . " is " .
$arr[$index]);
// This code is contributed by Ajit.
?>
PHP
$arr[$high])
return -1;
/* get the index of middle
element of arr[low..high] */
// low + (high - low)/2
$mid = ($low + $high)/2;
/* If x is same as middle element,
then return mid */
if($arr[$mid] == $x)
return $mid;
/* If x is greater than arr[mid],
then either arr[mid + 1] is
ceiling of x or ceiling lies
in arr[mid+1...high] */
else if($arr[$mid] < $x)
{
if($mid + 1 <= $high &&
$x <= $arr[$mid + 1])
return $mid + 1;
else
return ceilSearch($arr, $mid + 1,
$high, $x);
}
/* If x is smaller than arr[mid],
then either arr[mid] is ceiling
of x or ceiling lies in
arr[low....mid-1] */
else
{
if($mid - 1 >= $low &&
$x > $arr[$mid - 1])
return $mid;
else
return ceilSearch($arr, $low,
$mid - 1, $x);
}
}
// Driver Code
$arr = array(1, 2, 8, 10, 10, 12, 19);
$n = sizeof($arr);
$x = 20;
$index = ceilSearch($arr, 0, $n - 1, $x);
if($index == -1)
echo("Ceiling of $x doesn't exist in array ");
else
echo("ceiling of $x is");
echo(isset($arr[$index]));
// This code is contributed by nitin mittal.
?>
输出 :
ceiling of 3 is 8
时间复杂度: O(n)
方法 2(二分查找)
这里没有使用线性搜索,而是使用二进制搜索来查找索引。二分搜索将时间复杂度降低到 O(Logn)。
PHP
$arr[$high])
return -1;
/* get the index of middle
element of arr[low..high] */
// low + (high - low)/2
$mid = ($low + $high)/2;
/* If x is same as middle element,
then return mid */
if($arr[$mid] == $x)
return $mid;
/* If x is greater than arr[mid],
then either arr[mid + 1] is
ceiling of x or ceiling lies
in arr[mid+1...high] */
else if($arr[$mid] < $x)
{
if($mid + 1 <= $high &&
$x <= $arr[$mid + 1])
return $mid + 1;
else
return ceilSearch($arr, $mid + 1,
$high, $x);
}
/* If x is smaller than arr[mid],
then either arr[mid] is ceiling
of x or ceiling lies in
arr[low....mid-1] */
else
{
if($mid - 1 >= $low &&
$x > $arr[$mid - 1])
return $mid;
else
return ceilSearch($arr, $low,
$mid - 1, $x);
}
}
// Driver Code
$arr = array(1, 2, 8, 10, 10, 12, 19);
$n = sizeof($arr);
$x = 20;
$index = ceilSearch($arr, 0, $n - 1, $x);
if($index == -1)
echo("Ceiling of $x doesn't exist in array ");
else
echo("ceiling of $x is");
echo(isset($arr[$index]));
// This code is contributed by nitin mittal.
?>
输出 :
Ceiling of 20 doesn't exist in array
时间复杂度:O(Logn)
有关更多详细信息,请参阅排序数组中有关天花板的完整文章!