基于级别的嵌套数组

0: {content: "Heading 1 2 3 4 5", level: 2, anchor: "heading-1-2-3-4-5", className: "testtest", fontWeight: "", …}
1: {content: "Heading 2", level: 2, anchor: "heading-2", fontWeight: "", textTransform: "", …}
2: {content: "Inner Heading", level: 2, anchor: "inner-heading", fontWeight: "", textTransform: "", …}
3: {content: "Heading Level 3", level: 3, anchor: "heading-level-3", fontWeight: "", textTransform: "", …}
4: {content: "Heading Level 3-2", level: 3, fontWeight: "", textTransform: "", noBottomSpacing: false, …}
5: {content: "Heading Level 4", level: 4, anchor: "heading-level-4", fontWeight: "", textTransform: "", …}
6: {content: "Heading Level 2", level: 2, anchor: "heading-level-2", fontWeight: "", textTransform: "", …}
7: {content: "Heading 4", level: 6, anchor: "heading-4", fontWeight: "", textTransform: "", …}

我在 JavaScript 中有这个数组，我需要根据级别嵌套它。

例如，

level 2
level 2
level 2
   level 3
   level 3
     level 4
level 2
  level 4

嵌套这个的最好方法是什么？

到目前为止，我已经尝试过：

        const nestHeading = heading => {
            const nestedHeadersLength = nestedHeaders.length;

            if ( nestedHeadersLength >= 1 ) {
                const previousIndex = nestedHeadersLength - 1;

                if ( previousIndex >= 0 ) {
                    if (
                        heading.level !== nestedHeaders[ previousIndex ].level &&
                        heading.level > nestedHeaders[ previousIndex ].level
                    ) {
                        nestedHeaders[ previousIndex ].innerHeadings.push( heading );
                    } else {
                        nestedHeaders.push( heading );
                    }
                }
            } else {
                nestedHeaders.push( heading );
            }
        };

标题是原始数据集中的每个项目。

您可以为级别采用一个辅助数组，并为每个级别的索引采用一个辅助数组，作为对非零基或缺失级别的调整。

想象一下，你所有的level属性从零开始运行，并建立在该值之上并且没有漏洞，您只能采用该行

levels[index].push({ ...o, children: levels[index + 1] = [] });
//     ^^^^^                                                      parent
//                                          ^^^^^^^^^             children

with o.level代替index.

那么只有levels用于将节点移动到树的正确位置。每个级别都取决于之前的（较小的）级别。

级别不从零开始以及缺少级别的主要问题是将级别调整为有意义的索引。这是通过使用级别作为值并使用这些值的数组的索引来存档的。获取索引的规则是，如果没有找到，则取数组的最后一个长度，并将级别推送到indices大批。否则短路indices数组到length of index加一，以防止在数组中找到更深的嵌套符号。

var data = [{ content: "Heading 1 2 3 4 5", level: 2 }, { content: "Heading 2", level: 2 }, { content: "Inner Heading", level: 2 }, { content: "Heading Level 3", level: 3 }, { content: "Heading Level 3-2", level: 3 }, { content: "Heading Level 4", level: 4 }, { content: "Heading Level 2", level: 2 }, { content: "Heading 4", level: 6 },  { content: "Heading 1", level: 1 }, { content: "Heading 5", level: 5 }],
    result = [],
    indices = [],
    levels = [result];
   
data.forEach(o => {
    var index = indices.findIndex(level => level >= o.level);
    if (index === -1) {
        index = indices.push(o.level) - 1;
    } else {
        indices.length = index + 1;
    }
    levels[index].push(Object.assign({}, o, { children: levels[index + 1] = [] }));
});

console.log(result);

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