我已经有解决方案了。但我认为它会更加优化。所以请为我提供一个解决方案。并记住不要使用 php 的预定义函数。就像 max() 函数一样。 我知道有很多方法可以找到它,但我想要最好的解决方案。因为我的数组包含超过 10 万条记录,并且需要花费大量时间。或者有时网站会关闭。
我的代码:
<?php
$array = array('1', '15', '2','10',4);
echo "<pre>";
print_r($array);
echo "<pre>";
$max = 0;
$s_max=0;
for($i=0; $i<count($array); $i++)
{
$a = $array[$i];
$tmax = $max;
$ts_max = $s_max;
if($a > $tmax && $a > $ts_max)
{
$max = $a;
if($tmax > $ts_max) {
$s_max = $tmax;
} else {
$s_max = $ts_max;
}
} else if($tmax > $a && $tmax > $ts_max)
{
$max = $tmax;
if($a > $ts_max) {
$s_max = $a;
} else {
$s_max = $ts_max;
}
} else if($ts_max > $a && $ts_max > $tmax)
{
$max = $ts_max;
if($a > $tmax)
{
$s_max = $a;
} else {
$s_max = $tmax;
}
}
}
echo "Max=".$max;
echo "<br />";
echo "S_max=".$s_max;
echo "<br />";
?>
<?php
$array = array('200', '15','69','122','50','201');
$max_1 = $max_2 = 0;
for ($i=0; $i<count($array); $i++) {
if ($array[$i] > $max_1) {
$max_2 = $max_1;
$max_1 = $array[$i];
} else if ($array[$i] > $max_2 && $array[$i] != $max_2) {
$max_2 = $array[$i];
}
}
echo "Max=".$max_1;
echo "<br />";
echo "Smax 2=".$max_2;
请参阅此解决方案。
<?php
$numbers = array_unique(array(1,15,2,10,4));
// rsort : sorts an array in reverse order (highest to lowest).
rsort($numbers);
echo 'Highest is -'.$numbers[0].', Second highest is -'.$numbers[1];
// O/P: Highest is -15, Second highest is -10
?>
我没有检查你的解决方案,但就复杂性而言,IMO 是最佳的。如果数组没有更多的结构信息(比如已排序),则无法跳过条目。 IE。最好的解决方案是 O(n),您的解决方案就是。
这是一个完美且最短的代码,用于从数组中找出第二大值。如果数组仅包含一个值,下面的代码将始终返回值。
Example 1.
$arr = [5, 8, 1, 9, 24, 14, 36, 25, 78, 15, 37];
asort($arr);
$secondLargestVal = $arr[count($arr)-1];
//this will return 37
Example 2.
$arr = [5];
asort($arr);
$secondLargestVal = $arr[count($arr)-1];
//this will return 5
您还可以使用冒泡排序等排序技术
function bubble_Sort($my_array )
{
do
{
$swapped = false;
for( $i = 0, $c = count( $my_array ) - 1; $i < $c; $i++ )
{
if( $my_array[$i] > $my_array[$i + 1] )
{
list( $my_array[$i + 1], $my_array[$i] ) =
array( $my_array[$i], $my_array[$i + 1] );
$swapped = true;
}
}
}
while( $swapped );
return $my_array;
}
$test_array = array(3, 0, 2, 5, -1, 4, 1);
echo "Original Array :\n";
echo implode(', ',$test_array );
echo "\nSorted Array\n:";
echo implode(', ',bubble_Sort($test_array)). PHP_EOL;
Original Array :
3, 0, 2, 5, -1, 4, 1
Sorted Array :
-1, 0, 1, 2, 3, 4, 5
$array = array(80,250,30,40,90,10,50,60,50); // 250 2-times
$max = $max2 = 0;
foreach ($array as $key =>$val) {
if ($max < $val) {
$max2 = $max;
$max = $val;
} elseif ($max2 < $val && $max != $val) {
$max2 = $val;
}
}
echo "Highest Value is : " . $max . "<br/>"; //output: 250
echo "Second highest value is : " . $max2 . "<br/>"; //output: 90
您可以尝试以下代码
<?php
$array = array('1', '15', '2','10', 4);
$max_number = $second_max = PHP_INT_MIN;
foreach ($array as $value) {
$number = (int)$value; // convert value into integer
if ($number > $max_number) {
$second_max = $max_number;
$max_number = $number;
} elseif ($number > $second_max && $number != $max_number) {
$second_max = $number;
}
}
echo "Max = $max_number";
echo "Second Max = $second_max";
?>
“Kanishka Panaldeniya”给出的答案对于最高值来说是可以的,但对于第二高值会失败,即如果数组有 2 个相似的最高值,那么它将显示相同的最高值和第二高值。我已经通过添加一个级别比较来整理它,并且效果很好。
$array = array(50,250,30,250,40,70,10,50); // 250 2-times
$max=$max2=0;
for ($i = 0; $i < count($array); $i++) {
if ($array[$i] > $max) {
$max2 = $max;
$max = $array[$i];
} else if (($array[$i] > $max2) && ($array[$i] != $max)) {
$max2 = $array[$i];
}
}
echo "Highest Value is : " . $max . "<br/>"; //output : 250
echo "Second highest value is : " . $max2 . "<br/>"; //output : 70
此代码将从数组返回第二个最大值
$array = array(80,250,30,250,40,90,10,50,60,50); // 250 2-times
$max=$max2=0;
for ($i = 0; $i < count($array); $i++) {
if($i == 0) {
$max2 = $array[$i];
}
if($array[$i] > $max) {
$max = $array[$i];
}
if($max > $array[$i] && $array[$i] > $max2) {
$max2 = $array[$i];
}
}
echo "Highest Value is : " . $max . "<br/>"; //output : 250
echo "Second highest value is : " . $max2 . "<br/>"; //output : 90
Two way find the second highest salary
1. Sort the data in Descending order
2. get array first value
3. Check the condition already comments
$array = [2,3,6,11,17,14,19];
$max = $array[0];
$count = count($array);
for($i=0; $i<$count;$i++)
{
for($j=$i+1;$j<$count;$j++)
{
if($array[$i] < $array[$j])
{
$temp = $array[$i];
$array[$i] = $array[$j];
$array[$j] = $temp;
}
}
}
First Method
//echo $array[1]; // Second highest value
$second ='';
for($k=0;$k<2;$k++)
{
if($array[$k] >$max)
{
$second = $array[$k];
}
}
echo $second; // Second method
不使用MAX功能。在这里。
$arr = [3,4,-5,-3,1,0,4,4,4];
rsort($arr); // SORT ARRAY IN DESCENDING ORDER
$largest = $arr[0]; // IN DESCENDING ORDER THE LARGEST ELEMENT IS ALWAYS THE FIRST ELEMENT
$secondLargest = 0;
foreach($arr as $val){
$secondLargest = $val; // KEEP UPDATING THE VARIABLE WITH THE VALUE UNTIL IT RECEIVES THE FIRST ELEMENT WHICH IS DIFFERENT FROM THE LARGEST VALUE
if($val != $arr[0]){
break; // BREAK OUT OF THE LOOP AS SOON AS THE VALUE IS DIFFERENT THAN THE LARGEST ELEMENT
}
}
echo $secondLargest;