这样写二分查找怎么不对呢

<php function binarySearch($arr,$a){        $low = 0;    $high = count($arr)-1;    $mid = ceil(($low+$high)/2);while($low$arr[$mid]){        $high = count($arr)-1;    	$low = $mid+1;	    $mid = ceil(($low+$high)/2);    }    if ($a

回复讨论(解决方案)

function binarySearch($arr,$a){

$low = 0;
$high = count($arr)-1;
）本文来（源gaodai#ma#com搞@@代~&码*网2

搞代gaodaima码

$mid = ceil(($low+$high)/2);
$n = 0;防止死循环的措施
while($low<=$high && $n++<count($arr)){
if($a==$arr[$mid]){
echo “查找的数的位置是第”.”$mid”;
break;
}

if($a>$arr[$mid]){

$high = count($arr)-1;
$low = $mid -1; //注意这里
$mid = ceil(($low+$high)/2);
}

if ($a<$arr[$mid]) {

$low = 0;
$high = $mid +1; //注意这里
$mid = ceil(($low+$high)/2);
}
}
}

按 php 实际可写作

function binarySearch($arr,$a){  $low = 0;  $high = count($arr)-1;  $mid = ceil(($low+$high)/2);  $num = count($arr);  while($low $arr[$mid] ? array($mid, $high) : array($low, $mid);    $mid = ceil(($low+$high)/2);  }}

本帖最后由 xuzuning 于 2013-03-30 17:42:33 编辑
function binarySearch($arr,$a){

$low = 0;
$high = count($arr)-1;
$mid = ceil(($low+$high)/2);
$n = 0;防止死循环的措……
我有点晕 如果是四个数呢，比如$r=array(1,2,3,4) 查找1 貌似不行 用floor来求$mid 这个 我 理解，用ceil貌似 有点问题、、、、

按 php 实际可写作
PHP code?1234567891011121314function binarySearch($arr,$a){ $low = 0; $high = count($arr)-1; $mid = ceil(($low+$high)/2); $num = count($arr); while($low<=$high &&amp……[/quote
还是要 谢谢 版主大大、、、

function binarySearch($arr,$a){  $low = 0;  $high = count($arr)-1;  $mid = ceil(($low+$high)/2);  $loop = count($arr);  while($low$arr[$mid] ? array($mid, $high) : array($low, $mid);    $mid = ceil(($low+$high)/2);  }  echo "查找的数的位置在第{$low}、{$high}之间";}$arr1 = array(5,7,9,10,12,16,19);binarySearch($arr1,11);

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