Subsets II

最新推荐文章于 2022-05-07 14:28:24 发布

最新推荐文章于 2022-05-07 14:28:24 发布 · 85 阅读

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本文介绍了一种解决含重复元素数组的所有可能子集问题的算法。该算法通过深度优先搜索实现，确保子集按非递减顺序排列且不包含重复集合。文章提供了详细的Java代码实现。

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

Note:

Elements in a subset must be in non-descending order.
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],
  []
]

public class Solution {
	public List<List<Integer>> subsetsWithDup(int[] nums) {
    	List<List<Integer>> res = new ArrayList<List<Integer>>();
    	if (nums.length<=0 || nums==null) {
    		return res;
    	}
    	Arrays.sort(nums);
    	ArrayList<Integer> list = new ArrayList<Integer>();
    	for (int i = 1; i <= nums.length; i++) {
    		dfs(nums, 0, i, list, res);
    	}
    	res.add(new ArrayList<Integer>());
    	return res;
    }

	private void dfs(int[] arr, int start, int len, ArrayList<Integer> list,
			List<List<Integer>> res) {
		if (list.size() == len) {
			if (!res.contains(list)) {
				res.add(new ArrayList<Integer>(list));
			}
			return;
		}
		for (int i = start; i < arr.length; i++) {
			list.add(arr[i]);
			dfs(arr, i+1, len, list, res);
			list.remove(list.size()-1);
		}
	}
}