javascript 树状搜寻 数组
前端上的需求,需要写一个搜寻的 function 来遍历这个树，
function search(nodes, keyword){
//预期输入 如下方的 `const nodes`
}
const searchedNodes = search(nodes, "1-1-2-1");
// 预期要输出包含 "1-1-2-1"的所有节点以及该节点的父节点
/*
searchedNodes 应该要相等于
[
{
value: "1-1",
children: [
{ value: "1-1-2", children:[
{
value: "1-1-2-1",
children: [
{ value: "1-1-2-1-1" }
]
}
] }
]
}
]
*/
const nodes = [
{
value: "1-1",
children: [
{ value: "1-1-1"},
{ value: "1-1-2", children:[
{
value: "1-1-2-1",
children: [
{ value: "1-1-2-1-1" },
{ value: "1-1-2-1-2" }
]
},
{
value: "1-1-2-2"
}
] }
]
},
{
value: "1-2",
children: [
{ value: "1-2-1"},
{ value: "1-2-2", children:[
{
value: "1-2-2-1",
children: [
{ value: "1-2-2-1-1" },
{ value: "1-2-2-1-2" }
]
},
{
value: "1-2-2-2"
}
] }
]
},
];
请问这样的树搜寻应该要怎麽完成呢？
昨晚自己折騰了一會 寫了一個 DFS 的版本 效能似乎不太好 但堪用
const search = (nodes, keyword) => {
let newNodes = [];
for (let n of nodes) {
if (n.children) {
const nextNodes = this.keywordFilter(n.children, keyword);
if (nextNodes.length > 0) {
n.children = nextNodes;
} else if (n.label.toLowerCase().includes(keyword.toLowerCase())) {
n.children = nextNodes.length > 0 ? nextNodes : [];
}
if (
nextNodes.length > 0 ||
n.label.toLowerCase().includes(keyword.toLowerCase())
) {
newNodes.push(n);
}
} else {
if (n.label.toLowerCase().includes(keyword.toLowerCase())) {
newNodes.push(n);
}
}
}
return newNodes;
};
繁华开满天机2回答
-
HUX布斯
大概写了一下：function compare(value, keyword){ if(value === keyword){ return 2; } if(new RegExp('^' + value).test(keyword)){ return 1; } return 0;}function walk(node, keyword){ let isFind = false; const path = []; let children = null; function dfs(node, keyword){ if(compare(node.value, keyword) === 1){ path.push(node.value); node.children && node.children.forEach(childNode => { dfs(childNode, keyword); }); } if(compare(node.value, keyword) === 2) { node.children && (children = node.children); isFind = true; } } dfs(node, keyword); return { isFind, path, children }}function generateObj(pathArr, keyword, children){ const resObj = {}; let tempObj = resObj; pathArr.forEach(path => { tempObj.value = path; tempObj.children = [{}]; tempObj = tempObj.children[0]; }); tempObj.value = keyword; children && (tempObj.children = children); return resObj;}function search(nodes, keyword){ const searchedNodes = []; nodes.forEach(node => { const { isFind = false, path, children = null } = walk(node, keyword); if(isFind){ searchedNodes.push(generateObj(path, keyword, children)); } }); return searchedNodes;}const nodes = [ { value: "1-1", children: [ { value: "1-1-1"}, { value: "1-1-2", children:[ { value: "1-1-2-1", children: [ { value: "1-1-2-1-1" }, { value: "1-1-2-1-2" } ] }, { value: "1-1-2-2" } ] } ] }, { value: "1-2", children: [ { value: "1-2-1"}, { value: "1-2-2", children:[ { value: "1-2-2-1", children: [ { value: "1-2-2-1-1" }, { value: "1-2-2-1-2" } ] }, { value: "1-2-2-2" } ] } ] },];const searchedNodes = search(nodes, "1-1-2-1"); -
呼唤远方
你给的期望数据和 问题描述有歧义啊， children: [ { value: "1-1-2-1-1" }, { value: "1-1-2-1-2" } ]中的{ value: "1-1-2-1-2" }也是符合1-1-2-1的啊
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