PHP程序从两个已排序的数组中找到最接近的对
给定两个已排序的数组和一个数字 x,找到总和最接近 x的对,并且该对具有每个数组中的一个元素。
给定两个数组 ar1[0…m-1] 和 ar2[0..n-1] 和一个数字 x,我们需要找到一对 ar1[i] + ar2[j] 使得 (ar1 [i] + ar2[j] – x) 是最小值。
例子:
Input: ar1[] = {1, 4, 5, 7};
ar2[] = {10, 20, 30, 40};
x = 32
Output: 1 and 30
Input: ar1[] = {1, 4, 5, 7};
ar2[] = {10, 20, 30, 40};
x = 50
Output: 7 and 40
PHP
= 0)
{
// If this pair is closer to
// x than the previously
// found closest, then
// update res_l, res_r and
// diff
if (abs($ar1[$l] + $ar2[$r] -
$x) < $diff)
{
$res_l = $l;
$res_r = $r;
$diff = abs($ar1[$l] +
$ar2[$r] - $x);
}
// If sum of this pair is
// more than x, move to smaller
// side
if ($ar1[$l] + $ar2[$r] > $x)
$r--;
// move to the greater side
else
$l++;
}
// Print the result
echo "The closest pair is [", $ar1[$res_l], ", ", $ar2[$res_r], "] \n";
}
// Driver Code
$ar1 = array(1, 4, 5, 7);
$ar2 = array(10, 20, 30, 40);
$m = count($ar1);
$n = count($ar2);
$x = 38;
printClosest($ar1, $ar2, $m, $n, $x);
// This code is contributed by anuj_67.
?>
输出:
The closest pair is [7, 30]
有关更多详细信息,请参阅有关从两个已排序数组中查找最近对的完整文章!