简单的PHP程序需要更少的时间来执行
我最近申请了一份工作,要求完成一项测试,然后面试 测试中给出了两个问题,非常简单,我成功地完成了测试,但仍然有人告诉我测试失败,因为脚本花了超过18秒完成执行。这是一个程序,我不明白我还能做些什么来加快进度。虽然我考试不及格,但还是想知道我还能做些什么? 程序语言是PHP,我必须使用命令行输入 问题是:简单的PHP程序需要更少的时间来执行,php,algorithm,std,Php,Algorithm,Std,我最近申请了一份工作,要求完成一项测试,然后面试 测试中给出了两个问题,非常简单,我成功地完成了测试,但仍然有人告诉我测试失败,因为脚本花了超过18秒完成执行。这是一个程序,我不明白我还能做些什么来加快进度。虽然我考试不及格,但还是想知道我还能做些什么? 程序语言是PHP,我必须使用命令行输入 问题是: K Difference Given N numbers , [N<=10^5] we need to count the total pairs of numbers that hav
K Difference
Given N numbers , [N<=10^5] we need to count the total pairs of numbers that have a difference of K. [K>0 and K<1e9]
Input Format:
1st line contains N & K (integers).
2nd line contains N numbers of the set. All the N numbers are assured to be distinct.
Output Format:
One integer saying the no of pairs of numbers that have a diff K.
Sample Input #00:
5 2
1 5 3 4 2
Sample Output #00:3
Sample Input #01:
10 1
363374326 364147530 61825163 1073065718 1281246024 1399469912 428047635 491595254 879792181 1069262793
Sample Output #01:
0
Note: Java/C# code should be in a class named "Solution"
Read input from STDIN and write output to STDOUT.
K差
给定N个数字,[N0和k0;$i--){
对于($j=$i-1;$j>=0;$j--){
如果($ari_nums[$i]-$ari_nums[$j]==$diff){
$count++;
$j=0;
}
}
}
fprintf($fw,“%d,$count);
算法的运行时间为O(N^2),约为10^5*10^5=10^10。通过一些基本观察,可以将其简化为O(NlgN),约为10^5*16=1.6*10^6
算法:
for($i=$total\u nums-1;$i>0;$i--){
$hi=$i-1;
$low=0;
而($hi>=$low){
$mid=($hi+$low)/2;
如果($ary\u nums[$mid]==$ary\u nums[$i]-$diff){
$count++;
打破
}
如果($ari_nums[$mid]我在技术面试中得到了同样的问题。我想知道我们是否在为同一家公司面试。:)
无论如何,以下是我(面试后)得出的答案:
我认为这与您计算的$mid
(不考虑奇数)有关
$fr = fopen("php://stdin", "r");
$fw = fopen("php://stdout", "w");
fscanf($fr, "%d", $total_nums);
fscanf($fr, "%d", $diff);
$ary_nums = array();
for ($i = 0; $i < $total_nums; $i++) {
fscanf($fr, "%d", $ary_nums[$i]);
}
$count = 0;
sort($ary_nums);
for ($i = $total_nums - 1; $i > 0; $i--) {
for ($j = $i - 1; $j >= 0; $j--) {
if ($ary_nums[$i] - $ary_nums[$j] == $diff) {
$count++;
$j = 0;
}
}
}
fprintf($fw, "%d", $count);
for ($i = $total_nums - 1; $i > 0; $i--) {
$hi = $i-1;
$low = 0;
while($hi>=$low){
$mid = ($hi+$low)/2;
if($ary_nums[$mid]==$ary_nums[$i]-$diff){
$count++;
break;
}
if($ary_nums[$mid]<$ary_nums[$i]-$diff){
$low = $mid+1;
}
else{
$hi = $mid-1;
}
}
}
}
// Insert code to get the input here
$count = 0;
sort ($arr);
for ($i = 0, $max = $N - 1; $i < $max; $i++)
{
$lower_limit = $i + 1;
$upper_limit = $max;
while ($lower_limit <= $upper_limit)
{
$midpoint = ceil (($lower_limit + $upper_limit) / 2);
$diff = $arr[$midpoint] - $arr[$i];
if ($diff == $K)
{
// Found it. Increment the count and break the inner loop.
$count++;
$lower_limit = $upper_limit + 1;
}
elseif ($diff < $K)
{
// Search to the right of midpoint
$lower_limit = $midpoint + 1;
}
else
{
// Search to the left of midpoint
$upper_limit = $midpoint - 1;
}
}
}
Enter the numbers N and K: 10 3
Enter numbers for the set: 1 2 3 4 5 6 7 8 9 10
Result: 6