題目

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

Example 1:

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

Example 2:

Input: nums = [0,1]
Output: [[0,1],[1,0]]

Example 3:

Input: nums = [1]
Output: [[1]]

思路

跟 #46 一樣的方法，但要多考慮重複的數字。

Solution

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permuteUnique = function(nums) {
    const res = []
    
    nums.sort()
    
    dfs(res, [], nums)
    
    return res
};

var dfs = function(res, arr, nums) {
    var len = nums.length
    var temp1 = null
    var temp2 = null
    
    if (!len) return res.push(arr)
    
    for (var i = 0; i < len; i++) {
        
        if (i > 0 && nums[i] == nums[i-1]) continue
        
        temp1 = Array.from(arr)
        temp1.push(nums[i])
        
        temp2 = Array.from(nums)
        temp2.splice(i, 1)
        
        dfs(res, temp1, temp2)
    }
}

Complexity

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