47. Permutations II

Problem

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

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

Solution (Java)

class Solution {
    private List<List<Integer>> ans;

    public List<List<Integer>> permuteUnique(int[] nums) {
        ans = new ArrayList<>();
        permute(nums, 0);
        return ans;
    }

    private void permute(int[] nums, int p) {
        if (p >= nums.length - 1) {
            List<Integer> t = new ArrayList<>(nums.length);
            for (int n : nums) {
                t.add(n);
            }
            ans.add(t);
            return;
        }
        permute(nums, p + 1);
        boolean[] used = new boolean[30];
        for (int i = p + 1; i < nums.length; i++) {
            if (nums[i] != nums[p] && !used[10 + nums[i]]) {
                used[10 + nums[i]] = true;
                swap(nums, p, i);
                permute(nums, p + 1);
                swap(nums, p, i);
            }
        }
    }

    private void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}

Solution (Javascript)

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

    nums.sort((a, b) => (a - b));
    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++) {
    if (nums[i] === nums[i - 1]) continue;

    tmp1 = Array.from(arr);
    tmp1.push(nums[i]);

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

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

Explain:

跟之前一题的差别是可能有重复的数字，那就先排序一下，遇到重复的数字跳过就好。

Complexity:

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