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Learn how to find whether an ordered array contains a certain value in PHP

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2020-08-11 16:25:592126browse

Learn how to find whether an ordered array contains a certain value in PHP

Question: For an ordered array, how to determine whether a given value exists in the array.

Idea: To determine whether it exists, the simplest way is to directly loop through the array and compare each value. But for ordered arrays, writing like this completely fails to take advantage of the "ordered" feature.

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All we use "binary search",

//有序数组为

$arr = array(2,5,66,87,954,1452,5865);

//查找值

$str = 1452;

//我们先定义 三个参数

$front = 0;//一个开始值下标

$end = count($arr) - 1;//一个结束值下标

$mid = intval(($front + $end) / 2);//中间值下标

1. For the first comparison, we directly determine whether the search value str is equal to the intermediate value mid, and if it is equal to directly return true;

2. If the search value str is greater than the intermediate value mid, it means the search value str It may be on the right side of the middle value, that is, the start value front needs to be reassigned = the middle value mid 1, and the end value end does not need to be changed. The middle value mid is the new start value and end value in turn;

3. If you are looking for a value If str is less than the middle value mid, it means that the search value str may be to the left of the middle value, that is, the start value does not need to be changed, and the end value end needs to be reassigned = middle value - 1, and the middle value mid is the start value and the new end value in turn;

-----As above, compare the incoming start value, end value and intermediate value. Once the start value is greater than the end value, it means that it is not found and the query ends. Otherwise, it returns that it has been found.

The specific code is as follows:

$str = 89;//查找值

$arr = [1,55,66,89,420];//有序数组

$ren = find($arr, $str);

echo &#39;<pre class="brush:php;toolbar:false">&#39;;

var_dump($ren);

function find($arr, $str){

 $front = 0;//开始下标

 $end = count($arr) - 1;//结束下标

 while($front <= $end){//结束值 大于 开始值 ,反之则退出

  $mid = intval(($front + $end) / 2);//中间值下标

  if($str == $arr[$mid]){

   return $mid;//存在直接返回值的下标

  }

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

   $front = $mid + 1;//在前面

  }

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

   $end = $mid - 1;//在后面

  }

 }

 return false;

}

Return result: 89 is the fourth element value subscript 3

int(3)

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