Question

说我有这个ids / categories的树：

我需要一个函数来返回基于输入id的所有叶子（结束节点）。但只有来自每个分支中给出的最深的ID的节点。

我不确定如何解释，所以这里举几个例子：

f([1])      returns [2,4,5]
f([1,2])    returns [2]
f([1,3])    returns [4]
f([1,2,3])  returns [2,4]
f([3])      returns [4]
f([4])      returns [4]
f[1,6])     returns [2,4,5,8,9]
f[11])      returns [13, 14]
f[10])      returns [13, 14]

我正在使用的树的结构如下：

array(
    [category] => Object(Category)
    [children] => array(
        array(
            [category] => Object(Category)
            [children] => array(
                ...
            )
            [category] => Object(Category)
            [children] => array(
                ...
            )
    )

如果这样做更容易，我也可以使用平面阵列：

array(
    array(id, parent_id),
    array(id, parent_id),
    etc..
)

经过几个小时或搜索和撕裂，我甚至不确定这是否有意义。我怎么能这样做？

更新
根据评论;这里有一些准备好复制/粘贴的代码，带有测试用例正如您所看到的，它在测试[4]和[10]上失败了。

/*
Test data (tree):
0
    1
        2
        3
            4
        5
    6
        7
            8
            9
    10
        11
            12
                13
                14

Test cases:
f([1])      returns [2,4,5]
f([1,2])    returns [2]
f([1,3])    returns [4]
f([1,2,3])  returns [2,4]
f([3])      returns [4]
f([4])      returns [4]
f([1,6])    returns [2,4,5,8,9]
f([11])     returns [13, 14]
f([10])     returns [13, 14]

*/

$test[] = array('in'=>[1],      'out' => [2,4,5]);
$test[] = array('in'=>[1,2],    'out' => [2]);
$test[] = array('in'=>[1,3],    'out' => [4]);
$test[] = array('in'=>[1,2,3],  'out' => [2,4]);
$test[] = array('in'=>[1,2,5],  'out' => [2,5]);
$test[] = array('in'=>[3],      'out' => [4]);
$test[] = array('in'=>[4],      'out' => [4]);
$test[] = array('in'=>[1,6],    'out' => [2,4,5,8,9]);
$test[] = array('in'=>[11],     'out' => [13, 14]);
$test[] = array('in'=>[10],     'out' => [13, 14]);

echo '<pre>';
foreach($test as $t) {
    echo 'input: ' . implode(',',$t['in']) .' '.PHP_EOL;
    $r = f($t['in']);
    echo 'output: ' . implode(',',$r) .' ';
    if($r == $t['out']) {
        echo '(ok)';
    }
    else {
        echo '(TEST FAIL)'.PHP_EOL;
        echo 'got: ' . implode(',',$r) .' '.PHP_EOL;
        echo 'expected: ' . implode(',',$t['out']) .' ';
    }
    echo PHP_EOL.PHP_EOL;
}




function f($ids) {
    $nodes = getEndNodes(getTree(), $ids);
    return array_map(function($a){return $a['id'];},$nodes);    
}

function getEndNodes($tree, $ids, $force=false) {
    $end_nodes = array();
    foreach($tree as $el) {

        if(!empty($el['children'])) {

            // if given ids is in some of these children, search only those
            $children = array();
            foreach($el['children'] as $child) {
                if(in_array($child['element']['id'], $ids)) { $children[] = $child; }
            }
            // if no children in ids, search all (normal search)
            if(!empty($children)) {
                $end_nodes = array_merge($end_nodes, getEndNodes($children, $ids));
            }
            // if this element is in the given ids, force search
            elseif(in_array($el['element']['id'], $ids)) {
                $end_nodes = array_merge($end_nodes, getEndNodes($el['children'], $ids, true));
            }
            // if this is a force search
            elseif($force) {
                $end_nodes = array_merge($end_nodes, getEndNodes($el['children'], $ids));
            }
            else {
                // ??
            }
        }
        else {
            $end_nodes[] = $el['element'];
        }

    }
    return $end_nodes;
}


function getList() {
    $list = array(
        array('id'=>1,'parent_id'=>0),
        array('id'=>2,'parent_id'=>1),
        array('id'=>3,'parent_id'=>1),
        array('id'=>4,'parent_id'=>3),
        array('id'=>5,'parent_id'=>1),
        array('id'=>6,'parent_id'=>0),
        array('id'=>7,'parent_id'=>6),
        array('id'=>8,'parent_id'=>7),
        array('id'=>9,'parent_id'=>7),
        array('id'=>10,'parent_id'=>0),
        array('id'=>11,'parent_id'=>10),
        array('id'=>12,'parent_id'=>11),
        array('id'=>13,'parent_id'=>12),
        array('id'=>14,'parent_id'=>12),
    );
    return $list;
}

function getTree() {
    $hash = array();
    $list = getList();
    foreach($list as $el) {
        $hash[$el['id']] = array('element' => $el);
    }
    foreach($hash as $id => &$node) {
        if ($parent_id = $node['element']['parent_id']) {
            $hash[$parent_id]['children'][] =& $node;
        }
        else {
            $tree[] =& $node;
        }
    }
    unset($node, $hash);
    return $tree;
}

输出

input: 1 
output: 2,4,5 (ok)

input: 1,2 
output: 2 (ok)

input: 1,3 
output: 4 (ok)

input: 1,2,3 
output: 2,4 (ok)

input: 1,2,5 
output: 2,5 (ok)

input: 3 
output: 4 (ok)

input: 4 
output:  (TEST FAIL)
got:  
expected: 4 

input: 1,6 
output: 2,4,5,8,9 (ok)

input: 11 
output: 13,14 (ok)

input: 10 
output:  (TEST FAIL)
got:  
expected: 13,14

Answer 1

又过了一天，我想我得到了它，它并不漂亮，但它的确有效。如果有人需要这个：

input: 1
output: 2,4,5 (ok)

input: 1,2
output: 2 (ok)

input: 1,3
output: 4 (ok)

input: 1,2,3
output: 2,4 (ok)

input: 1,2,5
output: 2,5 (ok)

input: 3
output: 4 (ok)

input: 4
output: 4 (ok)

input: 1,6
output: 2,4,5,8,9 (ok)

input: 11
output: 13,14 (ok)

input: 10
output: 13,14 (ok)

input: 1,6,13
output: 2,4,5,8,9,13 (ok)

input: 8,12,6
output: 8,13,14 (ok)

input: 5
output: 5 (ok)

input: 0
output:  (ok)

input: 15
output: 15 (ok)

input: 16
output: 17 (ok)

input: 15,16
output: 15,17 (ok)

input: 999
output:  (ok)

输出

public static List<Integer> determineSplits(String fileName) throws FileNotFoundException, IOException
{
   PdfDocument pdfDocument = new PdfDocument(new PdfReader(fileName));
   List<Integer> splitPages = new ArrayList<>();
   for(int i=1;i<=pdfDocument.getNumberOfPages();i++) {
       String pageTxt = PdfTextExtractor.getTextFromPage(pdfDocument.getPage(i));
       if(pageTxt.contains("LoremIpsum"))
       {
           splitPages.add(1);
       }
   }
   pdfDocument.close();
}

通过给定的父ID获取分支的最深树节点

1 个答案: