给定的数组有重复的元素,所以基本上,我想找到我搜索过的元素最后一次出现的index
。
$arr = array(2, 3, 4, 4, 5, 6, 4, 8);
$x = 4; // number to search
$low = 0;
$high = count($arr)-1;
// I want this function to return 6 which is the index of the last occurrence of 4, but instead I'm getting -1 .
function binary_search_last_occ($arr, $x, $low, $high)
{
while ($low <=$high)
{
$mid = floor(($low+$high)/2);
$result = -1;
if ($arr[$mid] == $x)
{
// we want to find last occurrence, so go for
// elements on the right side of the mid
$result = $mid;
$low = $mid+1;
}
else if($arr[$mid] > $x)
{
$high = $mid-1;
}
else
$low = $mid+1;
}
return $result;
}
echo binary_search_last_occ($arr, $x, $low, $high); // outputs -1 ?
我不确定,为什么我得到-1。有什么建议吗?
答案 0 :(得分:2)
我没有看到你的循环,但我认为这很难用来获得这样的功能
$arr = array(2, 3, 4, 4, 5, 6, 4, 8);
$result = [];
foreach($arr as $key => $val){
$result[$val][] = $key;
}
echo end($result[4]);//6
或者您可以简单地使用asort
功能和array_search
一样使用
$arr = array(2, 3, 4, 4, 5, 6, 4, 8);
asort($arr);
echo array_search(4,$arr);//6
答案 1 :(得分:2)
首先,对于二进制搜索,您的数组必须进行排序,如果您的数组未排序,您可以使用简单的方法,如
function binary_search_last_occ($arr, $x, $low, $high)
{
$last_occ = -1;
while ($low <=$high)
{
if($arr[$low] == $x)
$last_occ = $low;
$low++;
}
return $last_occ ;
}
答案 2 :(得分:1)
并在while()上面定义$result
以避免每次使用-1
覆盖。因此,您将结果显示为-1。
$result = -1;
while ($low <=$high)
{
$mid = floor(($low+$high)/2);
if ($arr[$mid] == $x)
{
// we want to find last occurrence, so go for
// elements on the right side of the mid
$result = $mid;
$low = $mid+1;
}
else if($arr[$mid] > $x)
{
$high = $mid-1;
}
else
$low = $mid+1;
}
return $result;