LeetCode Subsets II

最新推荐文章于 2025-09-11 19:19:55 发布

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本文介绍了一种使用深度优先搜索（DFS）解决含重复元素数组的所有可能子集问题的算法。通过对输入数组进行排序并利用递归方法，该算法能够有效地生成所有不包含重复子集的组合，并确保每个子集内的元素按非递减顺序排列。

Description:

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],
  []
]

Solution:

似乎前面有类似这种去不同子集的题目，DFS边形即可。

import java.util.*;

public class Solution {
	List<List<Integer>> list = new ArrayList<List<Integer>>();
	int[] nums;
	int n;

	public List<List<Integer>> subsetsWithDup(int[] nums) {
		Arrays.sort(nums);
		this.nums = nums;
		this.n = nums.length;

		ArrayList<Integer> currentList = new ArrayList<Integer>();
		dfs(0, currentList);
		return list;
	}

	void dfs(int index, ArrayList<Integer> currentList) {
		if (index == n) {
			list.add(new ArrayList<Integer>(currentList));
			return;
		}

		int limit = index + 1;
		for (; limit < n; limit++)
			if (nums[limit] != nums[index])
				break;

		dfs(limit, currentList);
		for (int i = index; i < limit; i++) {
			currentList.add(nums[i]);
			dfs(limit, currentList);
		}
		for (int i = index; i < limit; i++)
			currentList.remove(currentList.size() - 1);
	}
}