Subsets II Leetcode

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,2], a solution is:

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

 

这道题和前面的差不多，却因为typo debug了好久。。。= = 需注意start和i哪个是哪个。

去重的过程和之前一样。

public class Solution {
    List<List<Integer>> result;
    List<Integer> current;
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        result = new ArrayList<>();
        current = new ArrayList<>();
        if (nums == null) {
            return result;
        }
        Arrays.sort(nums);
        helper(nums, 0);
        return result;
    }
    public void helper(int[] nums, int start) {
        result.add(new ArrayList<>(current));
        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i - 1]) {
                continue;
            }
            current.add(nums[i]);
            helper(nums, i + 1);
            current.remove(current.size() - 1);
        }
    }
}

这道题还有非递归的方法，比dfs稍微快一些。就是对于每个值，往现有的子集里加，形成新的子集。如果这个数与前面的数一样的话，就只往上一个数新生成的子集里面加就可以了。

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> current = new ArrayList<>();
        if (nums == null) {
            return result;
        }
        result.add(current);
        Arrays.sort(nums);
        int size = 0;
        int begin = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                begin = size;
            } 
            size = result.size();
            for (int j = begin; j < size; j++) {
                List<Integer> l = new ArrayList<>(result.get(j));
                l.add(nums[i]);
                result.add(l);
            }
            begin = 0;
        }
        return result;
    }
}