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

/**
 * @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 :