46. Permutations

Problem

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

Solution (Java)

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        if (nums == null || nums.length == 0) {
            return new ArrayList<>();
        }
        List<List<Integer>> finalResult = new ArrayList<>();
        permuteRecur(nums, finalResult, new ArrayList<>(), new boolean[nums.length]);
        return finalResult;
    }

    private void permuteRecur(
            int[] nums, List<List<Integer>> finalResult, List<Integer> currResult, boolean[] used) {
        if (currResult.size() == nums.length) {
            finalResult.add(new ArrayList<>(currResult));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (used[i]) {
                continue;
            }
            currResult.add(nums[i]);
            used[i] = true;
            permuteRecur(nums, finalResult, currResult, used);
            used[i] = false;
            currResult.remove(currResult.size() - 1);
        }
    }
}

Solution (Javascript)

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permute = function(nums) {
  var res = [];

  dfs(res, [], nums);

  return res;
};

var dfs = function (res, arr, nums) {
  var len = nums.length;
  var tmp1 = null;
  var tmp2 = null;

  if (!len) return res.push(arr);

  for (var i = 0; i < len; i++) {
    tmp1 = Array.from(arr);
    tmp1.push(nums[i]);

    tmp2 = Array.from(nums);
    tmp2.splice(i, 1);

    dfs(res, tmp1, tmp2);
  }
};

Explain:

nope.

Complexity:

• Time complexity : O(n!).
• Space complexity :