【leetcode】47. Permutations II

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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]
]

思考：可能会出现重复的数字，对于该类数字，我们无需对这两个数字进行交换，所以需要利用到Set对原始元素进行去重判断

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums.length <= 0){
            return res;
        }
        Arrays.sort(nums);
        dfs(res,nums,0);
        return res;
    }
    
    public void dfs(List<List<Integer>> res,int[] nums,int row){
        if(row == nums.length){
            List<Integer> temp = new ArrayList<>();
            for(Integer num:nums){
                temp.add(num);
            }
            res.add(temp);
            return;
        }
        Set<Integer> set = new HashSet<>();
        for(int i=row;i<nums.length;i++){
            if(set.add(nums[i])){
               swap(nums,i,row);
                dfs(res,nums,row+1);
                swap(nums,i,row); 
            }
            
        }
    }
    
    public void swap(int[] nums,int i,int j){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
        
    }
}