8个PHP数组面试题_php技巧
网上找的PHP数组题,准备自己做一遍并且记录下来。
1.写函数创建长度为10的数组,数组中的元素为递增的奇数,首项为1.
function arrsort($first,$length){
$arr = array();
for($i=$first;$i
$arr[] = $i*2-1;
}
return $arr;
}
$arr1 = arrsort(1,10);
print_r($arr1);
输出:
Array ( [0] => 1 [1] => 3 [2] => 5 [3] => 7 [4] => 9 [5] => 11 [6] => 13 [7] => 15 [8] => 17 [9] => 19 )
2.创建长度为10的数组,数组中的数为递增的等比数,比值为3,首项为1.
//$num为比值
function arrsort($first,$length,$num){
$arr= array();
for($i=$first;$i
//pow($num,$i-2);返回$num的($i-2)次方
$arr[] = $num*pow($num,$i-2);
}
return $arr;
}
$arr1 = arrsort(1,10,3);
print_r($arr1);
输出:
Array ( [0] => 1 [1] => 3 [2] => 9 [3] => 27 [4] => 81 [5] => 243 [6] => 729 [7] => 2187 [8] => 6561 [9] => 19683 )
3.求数组中最大数的下标.
function maxkey($arr){
$maxval = max($arr);
foreach($arr as $key=>$val){
if($maxval == $val){
$maxkey = $key;
}
}
return $maxkey;
}
$arr = array(0,-1,-2,5,"b"=>15,3);
echo maxkey($arr);
输出:
b
4.创建一个长度为10的数组,数组中的元素满足斐波拉契数列的规律.
( 斐波那契数列,又称黄金分割数列,指的是这样一个数列:1、1、2、3、5、8、13、21、……在数学上,斐波纳契数列以如下被以递归的方法定义:F0=0,F1=1,Fn=F(n-1)+F(n-2)(n>=2,n∈N*). 特别指出:第0项是0,第1项是第一个1。)
function arrFibo($len){
$arr[0] = 0;
$arr[1] = 1;
for($i=2;$i
$arr[$i] = $arr[$i-1]+$arr[$i-2];
}
return $arr;
}
echo "
";<br> print_r(arrFibo(10));<br> echo "
输出:
Array
(
[0] => 0
[1] => 1
[2] => 1
[3] => 2
[4] => 3
[5] => 5
[6] => 8
[7] => 13
[8] => 21
[9] => 34
)
5.计算数组中最大数和最小数的差.
两种方法:
①max/min
输出:
102
②sort把元素按从小到大排序/rsort吧元素按从大到小排序
function arrsub($arr){
sort($arr);
$min = $arr[0];
rsort($arr);
$max = $arr[0];
$sub = $max - $min;
return $sub;
}
$arr = array(-1,-2,100);
echo arrsub($arr);
输出:
102
6.写一个方法,将一个长度超过10的数组最后5项直接截取,不改变顺序变为前5项,如{1,2,3,4,5,6,7,8,9,10}变为{6,7,8,9,10,1,2,3,4,5}.
思路:先把数组截取相应的长度( array_slice ),再把2段数组拼接( array_merge )
function arrsort($arr){
$num = count($arr);
if($num > 10){
//array_slice($arr,起始位置,截取长度,保留索引(默认为false))
$arr_firstpart = array_slice($arr,0,$num-5,true);
$arr_lastpart = array_slice($arr,($num-5),5,true);
}else{
echo "数组不超过10个元素,请重新输入";
exit();
}
//拼接
$arr_new = array_merge($arr_lastpart,$arr_firstpart);
return $arr_new;
}
$arr = array("a"=>1,2,3,8,9,6,"b"=>5,-1,"c"=>8,0,7);
echo "
"; <p>print_r($arr);</p> <p>echo "<br>= = = = = 拼接后 = = = = <br><br>";</p> <p>print_r(arrsort($arr));</p> <p>echo "</p>
输出:
Array
(
[a] => 1
[0] => 2
[1] => 3
[2] => 8
[3] => 9
[4] => 6
[b] => 5
[5] => -1
[c] => 8
[6] => 0
[7] => 7
)
= = = = = 拼接后 = = = =
Array
(
[b] => 5
[0] => -1
[c] => 8
[1] => 0
[2] => 7
[a] => 1
[3] => 2
[4] => 3
[5] => 8
[6] => 9
[7] => 6
)
当数组不满足长度为10时:
$arr = array("a"=>1,2,3);
输出:
Array
(
[a] => 1
[0] => 2
[1] => 3
)
= = = = = 拼接后 = = = =
数组不超过10个元素,请重新输入
7.将两个数组连接成一个新数组.
方法①使用array_merge()函数
array_merge($arr1,$arr2);
方法②使用array_merge_recursive()函数递归追加数组
( array_merge_recursive() 函数与 array_merge() 函数 一样,将一个或多个数组的元素的合并起来,一个数组中的值附加在前一个数组的后面。并返回作为结果的数组。
但是,与 array_merge() 不同的是,当有重复的键名时,值不会被覆盖,而是将多个相同键名的值递归组成一个数组。)
$arr = array("a"=>1,"b"=>2,3);
$arr2 = array("a"=>Dee,3,5);
$arr3 = array_merge($arr,$arr2);
$arr4 = array_merge_recursive($arr,$arr2);
echo "
";<br> print_r($arr3); <p> echo "<br> = = = = = <br><br>";</p> <p> print_r($arr4);<br> echo "</p>
输出:
Array
(
[a] => Dee
[b] => 2
[0] => 3
[1] => 3
[2] => 5
)
= = = = =
Array
(
[a] => Array
(
[0] => 1
[1] => Dee
)
[b] => 2
[0] => 3
[1] => 3
[2] => 5
)
第一个输出的数组中索引为"a"的值1被索引被第二个数组中索引为"a"的值Dee覆盖了。
方法③
function arrsort($arr1,$arr2){
$arr_new = $arr1;
foreach($arr2 as $key=>$val){
$arr_new[] = $val;
}
return $arr_new;
}
$arr1 = array("a"=>1,"b"=>2,3);
$arr2 = array("a"=>Dee,"c"=>3,5);
echo "
";<br> print_r(arrsort($arr1,$arr2));<br> echo "
输出:
Array
(
[a] => 1
[b] => 2
[0] => 3
[1] => Dee
[2] => 3
[3] => 5
)
如果是索引数组而且有重复的索引,则第二个数组中的这个重复的索引会被修改成新的索引。
8.数组逆序( 不能使用rsort函数,不能生成新数组 )
使用array_reverse()函数会创建新的数组,所以不能使用。
$arr = array("a","b","c",1,10);
$i = "";//要替换位置的数的下标
$j = "";//临时变量
$k = "";//被替换位置的数的下标
$len = count($arr);
$half_len = floor($len/2);//向下取整,取整的值是循环的次数
for($i=0;$i
$j = $arr[$i];
//判断数组个数奇偶
if($len%2!=0){ //奇数
$k = $half_len*2-$i;
}else{
//偶数
$k = $half_len*2-$i-1;
}
$arr[$i] = $arr[$k];
$arr[$k] = $j;
}
echo "
";<br> print_r($arr);<br> echo "
输出:
Array
(
[0] => 10
[1] => 1
[2] => c
[3] => b
[4] => a
)

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