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leetcode 78. subsets

admin 11月 13, 2020 0

78. Subsets

Difficulty:: Medium

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

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Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

Solution

Language: Java

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class  {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null) {
            return result;
        }
        dfsHelper(nums, result, new ArrayList<Integer>(), 0);
        return result;
    }
    
    private void dfsHelper(int[] nums, List<List<Integer>> result, List<Integer> subset, int index) {
        result.add(new ArrayList<Integer>(subset));
        for (int i = index; i < nums.length; i++) {
            subset.add(nums[i]);
            dfsHelper(nums, result, subset, i + 1);
            subset.remove(subset.size() - 1);
        }
    }
}

使用位运算
一共有$2^n$种可能性，如果用一个32进制数来确定一个数是否存在，就直接遍历0到n位全1的数即可。这种情况建立在nums不会超过31个的基础上，因为数字个数多了的话子集数太大，所以一般不会超过这个数字

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class  {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null) {
            return result;
        }
        int len = nums.length;
        int max = 1;
        for (int i = 0; i < len - 1; i++) {
            max = (max << 1) + 1;
        }
        for (int i = 0; i <= max; i++) {
            List<Integer> subset = new ArrayList<>();
            int tmp = i;
            for (int j = 0; j < len; j++) {
                if ((tmp & 1) != 0) {
                    subset.add(nums[j]);
                }
                tmp >>= 1;
            }
            result.add(subset);
        }
        return result;
    }
    

}
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