Javascript：如何找到元素在多维数组中的位置

Question

I have a multi array like this:

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]]]

If i want search for " es22 ", how can i get the position like tree[3][3][0] ? I have tried like this:

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

Whenever the item is found, return the index , and append all previous indexes before it:

 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

This is my attempt...

 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')); 

Description:

A depth first search. If subtree is an array, then enumerate. For each element, run the depth first search on that subtree.

If subtree IS item return true . This will cause the stack to unwind. At each stack frame, if the subtree search was successful, add current array index to the front of path .

Pass path up the stack frame chain if item was found in the subtree.

When complete, if the element was found, path is returned containing indices to the element, otherwise false is returned.

Pseudocode:

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

Hope it helps:

The imeplemntation:

 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 []; } 

And the test:

 // 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) }