点击回首页
我的浏览记录 | | 帮助?
当前位置:
首页>> 行业软件>> 在线考试系统>> 源文件浏览
[免费版 Free] WebForm,下载次数:418 次 | 关键字: 考试 办公 高校 竞赛 企事业单位

源码截图

源码目录树

;
当前路径:vendor/sebastian/diff/src/MemoryEfficientLongestCommonSubsequenceCalculator.php
<?php declare(strict_types=1);
/*
 * This file is part of sebastian/diff.
 *
 * (c) Sebastian Bergmann <sebastian@phpunit.de>
 *
 * For the full copyright and license information, please view the LICENSE
 * file that was distributed with this source code.
 */

namespace SebastianBergmann\Diff;

final class MemoryEfficientLongestCommonSubsequenceCalculator implements LongestCommonSubsequenceCalculator
{
    /**
     * {@inheritdoc}
     */
    public function calculate(array $from, array $to): array
    {
        $cFrom = \count($from);
        $cTo   = \count($to);

        if ($cFrom === 0) {
            return [];
        }

        if ($cFrom === 1) {
            if (\in_array($from[0], $to, true)) {
                return [$from[0]];
            }

            return [];
        }

        $i         = (int) ($cFrom / 2);
        $fromStart = \array_slice($from, 0, $i);
        $fromEnd   = \array_slice($from, $i);
        $llB       = $this->length($fromStart, $to);
        $llE       = $this->length(\array_reverse($fromEnd), \array_reverse($to));
        $jMax      = 0;
        $max       = 0;

        for ($j = 0; $j <= $cTo; $j++) {
            $m = $llB[$j] + $llE[$cTo - $j];

            if ($m >= $max) {
                $max  = $m;
                $jMax = $j;
            }
        }

        $toStart = \array_slice($to, 0, $jMax);
        $toEnd   = \array_slice($to, $jMax);

        return \array_merge(
            $this->calculate($fromStart, $toStart),
            $this->calculate($fromEnd, $toEnd)
        );
    }

    private function length(array $from, array $to): array
    {
        $current = \array_fill(0, \count($to) + 1, 0);
        $cFrom   = \count($from);
        $cTo     = \count($to);

        for ($i = 0; $i < $cFrom; $i++) {
            $prev = $current;

            for ($j = 0; $j < $cTo; $j++) {
                if ($from[$i] === $to[$j]) {
                    $current[$j + 1] = $prev[$j] + 1;
                } else {
                    $current[$j + 1] = \max($current[$j], $prev[$j + 1]);
                }
            }
        }

        return $current;
    }
}
关于我们 | 顾问团队 | 发展历程 | 联系我们 | 源码上传
联系电话(Tel):4008-010-151(免长途)
地址:北京市海淀区大恒科技大厦五层 邮编:100080
Floor 5th,Daheng Building,Zhongguancun,Beijing,China,100080
51Aspx.com 版权所有 CopyRight © 2006-2023. 京ICP备09089570号 | 京公网安备11010702000869号