Question

link

Stats

Frequency	4
Difficulty	3
Adjusted Difficulty	4
Time to use	--------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Solution

This is a very classic question.

One solution is to recursively push elements into a list, until all elements are pushed. Make use of a ‘visited’ array.

Another solution is using 2 nested for-loops to keep adding next elements. Read this post:

Loop through the array, in each iteration, a new number is added to different locations of results of previous iteration. Start from an empty List.

Keep in mind: this type of permutation questions always requires insert/delete before/after a recursion.

In case you are interested, there is a third solution, swap element method. It is also explained in the same post:

Swap each element with each element after it.

My code

Adding elements one by one (recommended)

public class Solution {
    public List<List<Integer>> permute(int[] num) {
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        if (num == null || num.length == 0) {
            return ans;
        }
        int len = num.length;
        Arrays.sort(num);
        helper(ans, new ArrayList<Integer>(), num, len, new boolean[len]);
        return ans;
    }

    private void helper(List<List<Integer>> ans, List<Integer> path, int[] num, int len, boolean[] visited) {
        if (path.size() == len) {
            ans.add(new ArrayList<Integer>(path));
            return;
        }
        for (int i = 0; i < len; i++) {
            if (!visited[i]) {
                path.add(num[i]);
                visited[i] = true;
                helper(ans, path, num, len, visited);
                path.remove(path.size() - 1);
                visited[i] = false;
            }
        }
    }
}

The double nested for-loop method

public class Solution {
    public ArrayList<ArrayList<Integer>> permute(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

        //start from an empty list
        result.add(new ArrayList<Integer>());

        for (int i = 0; i < num.length; i++) {
            //list of list in current iteration of the array num
            ArrayList<ArrayList<Integer>> current = new ArrayList<ArrayList<Integer>>();

            for (ArrayList<Integer> l : result) {
                // # of locations to insert is largest index + 1
                for (int j = 0; j < l.size()+1; j++) {
                    // + add num[i] to different locations
                    l.add(j, num[i]);
                    ArrayList<Integer> temp = new ArrayList<Integer>(l);
                    current.add(temp);
                    l.remove(j);
                }
            }
            result = new ArrayList<ArrayList<Integer>>(current);
        }
        return result;
    }
}

The swap element method

public class Solution {
    public ArrayList<ArrayList<Integer>> permute(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        permute(num, 0, result);
        return result;
    }

    void permute(int[] num, int start, ArrayList<ArrayList<Integer>> result) {

        if (start >= num.length) {
            ArrayList<Integer> item = convertArrayToList(num);
            result.add(item);
        }

        for (int j = start; j <= num.length - 1; j++) {
            swap(num, start, j);
            permute(num, start + 1, result);
            swap(num, start, j);
        }
    }

    private ArrayList<Integer> convertArrayToList(int[] num) {
        ArrayList<Integer> item = new ArrayList<Integer>();
        for (int h = 0; h < num.length; h++) {
            item.add(num[h]);
        }
        return item;
    }

    private void swap(int[] a, int i, int j) {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }
}