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

Note: The solution set must not contain duplicate subsets.

Example:

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

难度:medium

题目:给定一不含重复元素的集合,求其所有子集。
注意:答案不能包含重复的子集。

思路:递归

Runtime: 1 ms, faster than 100.00% of Java online submissions for Subsets.
Memory Usage: 37.2 MB, less than 0.96% of Java online submissions for Subsets.

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            subsets(nums, 0, i + 1, new Stack<>(), result);
        }
        result.add(new ArrayList<>());
        
        return result;
    }
    
    private void subsets(int[] nums, int idx, int k, Stack<Integer> stack, List<List<Integer>> result) {
        if (k <= 0) {
            result.add(new ArrayList<>(stack));
            return;
        }
        for (int i = idx; i < nums.length - k + 1; i++) {
            stack.push(nums[i]);
            subsets(nums, i + 1, k - 1, stack, result);
            stack.pop();
        }
    }
}

linm
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