[LintCode] Permutations

Given a list of numbers, return all possible permutations.

You can assume that there is no duplicate numbers in the list.

Example

For nums = [1,2,3], the permutations are:

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

Challenge 

Do it without recursion.

 

Solution 1. Recursion

Highlighted line 21 decides at each level, which element should be picked first. Take [1,2,3] as an example, we know all permutations must start with 1, 2, or 3.

used flag ensures that we do not pick the same element more than once.

public class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> results = new ArrayList<>();
        if(nums == null || nums.length == 0) {
            results.add(new ArrayList<Integer>());
            return results;
        }
        boolean[] used = new boolean[nums.length];
        for(int i = 0; i < used.length; i++) {
            used[i] = false;
        }
        permuteDfs(results, new ArrayList<Integer>(), nums, used);
        return results;
    }
    private void permuteDfs(List<List<Integer>> results, List<Integer> list, 
                            int[] nums, boolean[] used) {
        if(list.size() == nums.length) {
            results.add(new ArrayList<Integer>(list));
            return;
        }
21         for(int i = 0; i < nums.length; i++) {
            if(used[i]) {
                continue;
            }
            list.add(nums[i]);
            used[i] = true;
            permuteDfs(results, list, nums, used);
            list.remove(list.size() - 1);
            used[i] = false;
        }
    }
}

 

Solution 2. Iteration

 

 

 

Related Problems

Print Numbers By Recursion

Permutation Sequence

Permutations II