Unexpected output of Javascript function

Question

I'm trying to write a function that can perform permutation.

For example, if I input [1, 2, 3] , the expected answer will be

[ [ 3, 2, 1 ], [ 3, 2, 1 ],[ 3, 2, 1 ],[ 3, 2, 1 ],[ 3, 2, 1 ],[ 3, 2, 1 ] ]

But instead of showing the answer, it returns [[ ],[ ],[ ],[ ],[ ]]

Any ideas?

 var permute = (nums) => { results = []; var backtrack = (nums, result) => { if (nums.length === result.length) { results.push(result); } else { for (var i = 0; i < nums.length; i++) { if (result.indexOf(nums[i]) > -1) { continue; } result.push(nums[i]); backtrack(nums, result); result.pop(); } } } backtrack(nums, []); return results; }; console.log(permute([1, 2, 3])); 

Answer 1

You could take a local copy of result by slicing this array to prevent the same object reference in the result set.

 var permute = (nums) => { var results = []; var backtrack = (nums, result) => { if (nums.length === result.length) { results.push(result.slice()); // push copy } else { for (var i = 0; i < nums.length; i++) { if (result.indexOf(nums[i]) > -1) { continue; } result.push(nums[i]); backtrack(nums, result); result.pop(); } } }; backtrack(nums, []); return results; }; console.log(permute([1, 2, 3]).map(a => a.join(' '))); 

A version without pushing and popping.

 var permute = (nums) => { var results = []; var backtrack = (nums, result) => { if (nums.length === result.length) { results.push(result); } else { for (var i = 0; i < nums.length; i++) { if (result.indexOf(nums[i]) > -1) { continue; } backtrack(nums, result.concat(nums[i])); // use a new array } } }; backtrack(nums, []); return results; }; console.log(permute([1, 2, 3]).map(a => a.join(' '))); 

Answer 2

Just Another approach using reverse(), an special case of permutation which OP gives as an example:

var arr = [ [ 3, 2, 1 ], [ 3, 2, 1 ],[ 3, 2, 1 ],[ 3, 2, 1 ],[ 3, 2, 1 ],[ 3, 2, 1 ] ];

var permuted = arr.map((num) => {

    return num.reverse();

});

console.log(permuted);