Javascript:如何找到元素在多维数组中的位置？

Question

我有一个这样的多个数组：

var tree= ["es1“、null、" es11”、null、"info for es11“、"param for es11”、"es12“、null、"info for es12”、param for es12]、["es2“、null、" es21”、null、"info for es21“、"param for es21”、"es22“、"info for es22”、"param for es22“、"param for es22]

如果我想搜索"es22"，我如何才能得到像tree[3][3][0]这样的职位？我试过这样做：

function recursion(obj,strs){
if (found) return;
for(var j=0;j<obj.length;j++){
    c++;
    if (isArray(obj[j])&&!found) {
        recursion(obj[j],strs);
    } else {
        if (obj[j]==strs&&!found) {
            arr=obj;
            found=true;
            return;
        }
    }
}

Answer 1

每当找到项目时，返回index，并在它之前追加所有以前的索引：

​

var tree=["root",null,["es1",null,["es11",null,"info for es11","param for es11"],["es12",null,"info for es12","param for es12"]],["es2",null,["es21",null,"info for es21","param for es21"],["es22",null,"info for es22","param for es22"]]];

function recursion(arr, str, indexes) {
  var result;

  for (var i = 0; i < arr.length; i++) {
    if (Array.isArray(arr[i])) {
      result = recursion(arr[i], str, indexes);
      
      if(result !== null) {
        return [i].concat(result);
      }
    } else if(arr[i] === str) {
      return i;
    }
  }
  
  return null;
}

console.log(recursion(tree, "es22", []));

​

Answer 2

这是我的尝试..。

​

var tree = [
    "root", 
    null, 
    ["es1", null, ["es11"], ["es12"]], 
    ["es2", null, ["es21"], ["es22"]]
];

const find = (subtree, item, path = []) => Array.isArray(subtree) 
    ? subtree.some((e, i) => (find(e, item, path) && path.unshift(i))) && path 
    : subtree === item;

console.log(find(tree, 'es22'));

​

描述：

深度优先搜索。如果subtree是数组，则枚举。对于每个元素，首先对该子树执行深度搜索。

如果subtree是item，则返回true。这将导致堆栈展开。在每个堆栈帧中，如果子树搜索成功，则将当前数组索引添加到path的前面。

如果在子树中找到path，则将item传递到堆栈帧链中。

当完成时，如果找到元素，则返回path，其中包含元素的索引，否则返回false。

伪码：

def solution(subtree, item, path)
  if subtree is not an array
    return subtree is item
  else
    for each index, value in subtree
      var found = solution(value, item, path)
      if found
        add index to path
        return path
      end if
    end for
  end if
end def

Answer 3

希望它有帮助：

：

function findPosition(search, neddle) {
    for (let i = 0; i < search.length; i++) {
        if (search[i] === neddle) {
            return [i];
        } else if (Array.isArray(search[i])) {
            const match = findPosition(search[i], neddle);
            if (match.length > 0) {
                return [i].concat(match);
            }
        }
    }
    return [];
}

​

和测试：

// findPosition TEST
const tree = [
    "root",
    null,
    [
        "es1",
        null,
        ["es11", null, "info for es11", "param for es11"],
        ["es12", null, "info for es12", "param for es12"]
    ],
    [
        "es2",
        null,
        ["es21", null, "info for es21", "param for es21"],
        ["es22", null, "info for es22", "param for es22"]
    ]
];


const expected = [3, 3, 0].join(',');
const actual = findPosition(tree, 'es22').join(',');

if (actual === expected) {
    console.log('pass');
} else {
    console.log('fail');
    console.log(actual, 'not equal to', expected)
}

​