Description

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

For example,
[1,1,2] have the following unique permutations:

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

Solution

DFS

注意這道題目里面可能有重復(fù)元素，不能用swap的方式，因?yàn)榻粨Q之后數(shù)組就不能保證有序，跳過(guò)duplicate的邏輯就會(huì)被打破。所以用一個(gè)輔助boolea[] used用來(lái)表示某個(gè)位置的元素是否被使用過(guò)。

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> permutations = new ArrayList<>();
        if (nums == null || nums.length < 1) {
            return permutations;
        }
        
        Arrays.sort(nums);
        // used[i] means whether num[i] has been used in current permutation
        boolean[] used = new boolean[nums.length];   
        List<Integer> permutation = new ArrayList<>();
        permuteRecur(nums, used, permutation, permutations);
        return permutations;
    }
    
    public void permuteRecur(int[] nums,
                            boolean[] used,
                            List<Integer> permutation,
                            List<List<Integer>> permutations) {
        if (permutation.size() == nums.length) {
            permutations.add(new ArrayList<>(permutation));
            return;
        }
        
        int i = 0;
        while (i < nums.length) {
            if (!used[i]) {
                permutation.add(nums[i]);
                used[i] = true;
                permuteRecur(nums, used, permutation, permutations);
                used[i] = false;
                permutation.remove(permutation.size() - 1);
                // find next different value
                while (++i < nums.length && nums[i - 1] == nums[i]) {}
            } else {
                ++i;
            }
        }
    }
}