Sure, here is a Python solution using recursion and backtracking:

```python
def permute(nums):
def backtrack(first = 0):
# if all integers are used up
if first == n:
output.append(nums[:])
for i in range(first, n):
# place i-th integer first
# in the current permutation
nums[first], nums[i] = nums[i], nums[first]
# use next integers to complete the permutations
backtrack(first + 1)
# backtrack
nums[first], nums[i] = nums[i], nums[first]

n = len(nums)
output = []
backtrack()
return output
```

Here's how it works:

1. If all integers are used up, add the permutation to the output.
2. Iterate over the integers from `first` to the end of array.
3. Place the integer at index `first` that we're iterating on at the beginning of the permutation, i.e. swap(nums[first], nums[i]).
4. Proceed to create all permutations which starts from `first` and keeps `first+1` in the remaining part.
5. Now we have to backtrack, so we swap the numbers back to restore the original array. We do this because we want to try a different possibility for the number at the `first` index.
6. Repeat the process until we have found all permutations.

Question

Sure, here is a Python solution using recursion and backtracking:

```python
def permute(nums):
    def backtrack(first = 0):
        # if all integers are used up
        if first == n:  
            output.append(nums[:])
        for i in range(first, n):
            # place i-th integer first 
            # in the current permutation
            nums[first], nums[i] = nums[i], nums[first]
            # use next integers to complete the permutations
            backtrack(first + 1)
            # backtrack
            nums[first], nums[i] = nums[i], nums[first]
            
    n = len(nums)
    output = []
    backtrack()
    return output
```

Here's how it works:

1. If all integers are used up, add the permutation to the output.
2. Iterate over the integers from `first` to the end of array.
3. Place the integer at index `first` that we're iterating on at the beginning of the permutation, i.e. swap(nums[first], nums[i]).
4. Proceed to create all permutations which starts from `first` and keeps `first+1` in the remaining part. 
5. Now we have to backtrack, so we swap the numbers back to restore the original array. We do this because we want to try a different possibility for the number at the `first` index.
6. Repeat the process until we have found all permutations.

Knowee AI · Accepted Answer

Sure, here is a Python solution using recursion and backtracking:

```python
def permute(nums):
    def backtrack(first = 0):
        # if all integers are used up
        if first == n:  
            output.append(nums[:])
        for i in range(first, n):
            # place i-th integer first 
            # in the current permutation
            nums[first], nums[i] = nums[i], nums[first]
            # use next integers to complete the permutations
            backtrack(first + 1)
            # backtrack
            nums[first], nums[i] = nums[i], nums[first]
            
    n = len(nums)
    output = []
    backtrack()
    return output
```

Here's how it works:

1. If all integers are used up, add the permutation to the output.
2. Iterate over the integers from `first` to the end of array.
3. Place the integer at index `first` that we're iterating on at the beginning of the permutation, i.e. swap(nums[first], nums[i]).
4. Proceed to create all permutations which starts from `first` and keeps `first+1` in the remaining part. 
5. Now we have to backtrack, so we swap the numbers back to restore the original array. We do this because we want to try a different possibility for the number at the `first` index.
6. Repeat the process until we have found all permutations.

Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

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