90. Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

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

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);  // 必须排序

        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path = new ArrayList<>();

        dfs(nums, 0, path, result);
        return result;
    }

    private static void dfs(int[] nums, int start, List<Integer> path,
                            List<List<Integer>> result) {
        result.add(new ArrayList<Integer>(path));

        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i-1]) continue;
            path.add(nums[i]);
            dfs(nums, i + 1, path, result);
            path.remove(path.size() - 1);
        }
    }
}

和subsets基本一样，在当前数字等于前一个数字的时候出现重复，需要跳过。

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        //回溯法
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        List<Integer> tmp = new ArrayList<Integer>();
        backtrack(result, tmp, nums, 0);
        return result;
    }
    private void backtrack(List<List<Integer>> result, List<Integer> tmp, int[] nums, int start){
        //add前需要进行判断result中是否已经包含该元素。有可能重复。
        if (!result.contains(tmp)){
            result.add(new ArrayList(tmp));
        }
        for (int i = start; i < nums.length; i++){
            tmp.add(nums[i]);
            backtrack(result, tmp, nums, i + 1);
            tmp.remove(tmp.size() - 1);
        }
    }
}

但是这样每次判断一下不是更容易理解吗。。。