Case description:
Write a PHP function. To find the largest continuous sum among any n positive and negative integers, the time complexity of the algorithm is required to be as low as possible;
For example: echo getMaxSum(array(-2, 1, 3, 9, -4, 2, 3, 5, - 3,-4,1,3));//The maximum continuous sum is (1,3,9,-4,2,3,5) The addition function returns 19
The code is as follows:
<?php header('content-type:text/html;charset=utf8 ');
//算法分析: //1、必须是整数序列 //2、如果整个序列不全是负数,最大子序列的第一项必须是正数, //否则最大子序列后面的数加起来再加上第一项的负数,其和肯定不是最大的; //3、如果整个序列都是负数,那么最大子序列的和是0; <pre name="code" class="html">//全负数序列很简单,不举例 $arr=array(-2,1,3,9,-4,2,3,5,-3,-4,1,3); $thissum=0; $maxsum=0; $start=0;//记录子序列的起始下标 $end=0;//记录子序列的结束下标 for($i=0;$i<count($arr);$i++){ $thissum+=$arr[$i];//取得当前子序列的和 if($thissum>$maxsum){//如果当前子序列的和大于当前最大子序列的和 $maxsum=$thissum;//改变当前最大子序列的和 $end=$i; }else if($thissum<0){//如果当前子序列的和小于0,则把下一个元素值假定为最大子序列的第一项,这里可以保证最大自序列的第一项一定是正数 $thissum=0;//前提这个序列不全是负数 $start=$i+1; } } $parr=array($start,$end,$maxsum); list($start,$end,$maxsum)=$parr; print_r($arr); echo '最大子序列是:'; for($i=$start;$i<=$end;$i++){ echo $arr[$i].' '; } echo '<br>'; echo '最大子序列的和是'.$maxsum; ?>
Array ( [0] => -2 [1] => 1 [2] => 3 [3] => 9 [4] => -4 [5] => 2 [6] => 3 [7] => 5 [8] => -3 [9] => -4 [10] => 1 [11] => 3 ) 最大子序列是:1 3 9 -4 2 3 5 <br>最大子序列的和是19
The above introduces the difference between direct free kick and indirect free kick. PHP finds the largest continuous sum among any n positive and negative integers, including the difference between direct free kick and indirect free kick. I hope friends are interested in PHP tutorials. Helps.