Subsets -------bit

最新推荐文章于 2025-01-23 23:46:20 发布

zjj2015

最新推荐文章于 2025-01-23 23:46:20 发布

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本文介绍了一种使用位操作生成整数集合所有可能子集的方法。针对输入的不同整数集合，该算法能返回所有可能的子集，并确保子集内元素按非递减顺序排列且不含重复子集。通过递归和迭代两种方式实现，适用于面试和技术挑战。

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Given a set of distinct integers, return all possible subsets.

Have you met this question in a real interview? Yes
Example
If S = [1,2,3], a solution is:

[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
Note
Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
Challenge
Can you do it in both recursively and iteratively?

    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        int n = nums.length;
        Arrays.sort(nums);

        // 1 << n is 2^n
        // each subset equals to an binary integer between 0 .. 2^n - 1
        // 0 -> 000 -> []
        // 1 -> 001 -> [1]
        // 2 -> 010 -> [2]
        // ..
        // 7 -> 111 -> [1,2,3]
        for (int i = 0; i < (1 << n); i++) {
            ArrayList<Integer> subset = new ArrayList<Integer>();
            for (int j = 0; j < n; j++) {
                // check whether the jth digit in i's binary representation is 1
                if ((i & (1 << j)) != 0) {
                    subset.add(nums[j]);
                }
            }
            result.add(subset);
        }

        return result;
    }
}