【问题标题】:Constructing a hierarchy tree in javascript在javascript中构建层次结构树
【发布时间】:2015-12-05 05:01:16
【问题描述】:

我正在尝试从SOLR 获得的方面为类别和子类别等构建层次结构树。输入格式为:

['445',
79,
'398',
73,
'710',
32,
'398|760',
28,
'398|760|779',
28,
'445|446',
10]

其中单引号中的数据代表类别,后面的数字代表频率。

给定上面的数组,我需要的输出格式应该是:

[
    {
        "id": 445,
        "count": 79,
        "children": [
            {
                "id": 446,
                "count": 10
            }
        ]
    },
    {
        "id": 398,
        "count": 73,
        "children": [
            {
                "id": 760,
                "count": 28,
                "children": [
                    {
                        "id": 779,
                        "count": 28
                    }
                ]
            }
        ]
    },
    {
        "id": 710,
        "count": 32
    }
]

我正在尝试用相同的方式构建树 - 以获得有效的解决方案,但无法做到这一点。有人知道如何让它工作——或任何其他时间有效的解决方案。

谢谢!

【问题讨论】:

  • 我对这里预期的 trie 目的感到相当困惑。通常,您会在 JavaScript 中创建一个 trie 用于自动完成目的,例如如果用户键入类别名称,显然您正在处理类别 id...您的目标是哪种类型的 trie 结构?给定398|730|607,我可以为您设计一棵树,例如{398:{freq:73,760:{freq:28,779:{freq:28}},730:{freq:18,607:{freq:18}}}},这就是您想要的吗?还是您想要一个基于类别名称的树,例如:{'j':{'a':{'v':{'a':{...}}}}}
  • 我已经编辑了问题以展示我正在努力的最终输出,我想使用 Trie 作为字典,然后使用 trie 构建最终的数据结构(基数树不是' t 仅用于自动完成)。但是,只要复杂性保持为 O(n),任何其他方法都可以工作。
  • 为什么不拆分数据先得到树的顶点,然后得到边呢?还是总是这样:你得到第一个边,然后是顶点?
  • 好吧,原始输入的来源超出了我的控制范围,所以我不能假设,但我当然可以修改它以获得你提到的格式 - 但我不明白这有什么帮助缓解复杂性。我已经通过在桶中递归地形成桶解决了这个问题——但正如在时间和空间方面的可怕权衡中所预料的那样!

标签: javascript arrays node.js trie


【解决方案1】:

这应该可以满足您的要求,但没有子节点的节点仍然具有 node.children 属性,因此您可以使用 node.children.length 查看树中任何节点有多少子节点

var o=["445", 79, "398", 73, "710", 32, "398|760", 28, "398|760|779", 28, "445|446", 26, "710|1045", 25, "445|452", 24, "381", 19, "445|943", 19, "398|730", 18, "398|730|607", 18, "367", 16, "445|446|451", 15, "351", 14, "351|363", 14, "351|363|365", 14, "381|395", 14, "381|395|566", 14, "445|526", 14, "445|526|769", 14, "367|372", 12, "710|1045|1119", 11, "398|410", 10, "398|483", 9, "445|452|743", 8, "367|372|377", 7, "398|483|757", 7, "445|446|792", 7, "445|452|744", 7, "445|452|719", 6, "398|410|411", 5];

var nodeMap={};
var nodeLevels=[];
for(var i=0;i<o.length;i+=2)
{
    var catLineage=o[i].split('|');
    var cat=catLineage[catLineage.length-1];
    var depth=catLineage.length;
    while(depth>nodeLevels.length){nodeLevels.push([]);}
    nodeMap[cat]={id:cat,count:o[i+1],depth:depth,parents:catLineage.slice(0,catLineage.length-1)};
    nodeLevels[depth-1].push(cat);
}
var tree=[];
var treeNodeLookup={};
for(var i=0;i<nodeLevels.length;i++)
{
    for(var j=0;j<nodeLevels[i].length;j++)
    {
        var nodeId=nodeLevels[i][j];
        var nodeDepth=nodeMap[nodeId].depth;
        var nodeCount=nodeMap[nodeId].count;
        var parents=nodeMap[nodeId].parents;
        var pointer={children:tree};
        if(parents.length>0){pointer=treeNodeLookup[parents[0]];}
        var node={id:nodeId,count:nodeCount,children:[]};
        pointer.children.push(node);
        treeNodeLookup[nodeId]=pointer.children[pointer.children.length-1];
    }
}
console.log(tree);

使用console.log(JSON.stringify(tree)); 我的输出是:

[{"id":"445","count":79,"children":[{"id":"446","count":26,"children":[]},{"id":"452","count":24,"children":[]},{"id":"943","count":19,"children":[]},{"id":"526","count":14,"children":[]},{"id":"451","count":15,"children":[]},{"id":"769","count":14,"children":[]},{"id":"743","count":8,"children":[]},{"id":"792","count":7,"children":[]},{"id":"744","count":7,"children":[]},{"id":"719","count":6,"children":[]}]},{"id":"398","count":73,"children":[{"id":"760","count":28,"children":[]},{"id":"730","count":18,"children":[]},{"id":"410","count":10,"children":[]},{"id":"483","count":9,"children":[]},{"id":"779","count":28,"children":[]},{"id":"607","count":18,"children":[]},{"id":"757","count":7,"children":[]},{"id":"411","count":5,"children":[]}]},{"id":"710","count":32,"children":[{"id":"1045","count":25,"children":[]},{"id":"1119","count":11,"children":[]}]},{"id":"381","count":19,"children":[{"id":"395","count":14,"children":[]},{"id":"566","count":14,"children":[]}]},{"id":"367","count":16,"children":[{"id":"372","count":12,"children":[]},{"id":"377","count":7,"children":[]}]},{"id":"351","count":14,"children":[{"id":"363","count":14,"children":[]},{"id":"365","count":14,"children":[]}]}]

【讨论】:

  • 非常感谢!看起来很完美——空的儿童财产根本不是问题!非常感谢您的帮助!我想没有办法将其归结为 O(n)!
  • 如果你想添加大 O 复杂度支持,你应该使用节点树,这与 trie 非常不同。我不认为像 JavaScript 这样的高级语言(由众多浏览器中的一个解释)或用于 node.js 的 JavaScript 引擎能够对这种性能计算有很大意义。但我可以理解,由于强类型化,Java 等低级语言将如何从中受益匪浅。
猜你喜欢
  • 1970-01-01
  • 2017-04-18
  • 1970-01-01
  • 1970-01-01
  • 2023-03-21
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
相关资源
最近更新 更多