求指教一个区间查值的问题
if($bbb>0&&$bbb<=100){$baifen=100;} elseif($bbb>100&&$bbb<=500){$baifen=500;} elseif($bbb>500&&$bbb<=1000){$baifen=1000;} elseif($bbb>1000&&$bbb<=5000){$baifen=5000;}
要实现这样一个功能
想简化写法 然后去网上搜到了二分法非递归的 但自己改了下 没有改成功
$arr[0,100,500,1000,5000];$b[100,500,1000,5000];
小于0的不考虑
想实现 如果$val=100 结果是$b[0];
有没有大神教我下用二分法非递归的 怎么改 实现这个
这个是网上找的原始的代码
function binarySearch($array,$value){ //起始位置,你可以作为函数参数 $low = 0; $high = count($array)-1; while($low <= $high){ $mid = ceil(($low+$high)/2); if($array[$mid] == $value) return $mid; if($array[$mid] < $value){ $low = $mid + 1; } if($array[$mid] > $value){ $high = $mid - 1; } } return -1; }
我修改了下 无法实现 感谢
回复讨论(解决方案)
直接条件比较
if($bbb > 1000) ) $baifen = 5000;elseif($bbb > 500) $baifen = 1000;elseif($bbb > 100) $baifen = 500;elseif($bbb > 0) $baifen = 100;
使用字典
$dict = array(0 => 100, 100 => 500, 500 => 1000, 1000 => 5000);krsort($dict);foreach($$dict as $k=>$v) if($x > $k) break;echo $v;
直接条件比较
if($bbb > 1000) ) $baifen = 5000;elseif($bbb > 500) $baifen = 1000;elseif($bbb > 100) $baifen = 500;elseif($bbb > 0) $baifen = 100;
使用字典
$dict = array(0 => 100, 100 => 500, 500 => 1000, 1000 => 5000);krsort($dict);foreach($$dict as $k=>$v) if($x > $k) break;echo $v;
感谢 。。 请教一下 返回$v 是得到那个最大值 比如 $x=101 $v返回的是500 。。。 我想返回的是1 因为他是在第二个范围内 ,我是取第一范围得到什么值 第二范围得到什么值~~ 该如何返回 感谢版主~~

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