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],
  []
]

簡便方法如下：

While iterating through all numbers, for each new number, we can either pick it or not pick it
1, if pick, just add current number to every existing subset.
2, if not pick, just leave all existing subsets as they are.
We just combine both into our result.

For example, {1,2,3} intially we have an emtpy set as result [ [ ] ]
Considering 1, if not use it, still [ ], if use 1, add it to [ ], so we have [1] now
Combine them, now we have [ [ ], [1] ] as all possible subset

Next considering 2, if not use it, we still have [ [ ], [1] ], if use 2, just add 2 to each previous subset, we have [2], [1,2]
Combine them, now we have [ [ ], [1], [2], [1,2] ]

Next considering 3, if not use it, we still have [ [ ], [1], [2], [1,2] ], if use 3, just add 3 to each previous subset, we have [ [3], [1,3], [2,3], [1,2,3] ]
Combine them, now we have [ [ ], [1], [2], [1,2], [3], [1,3], [2,3], [1,2,3] ]

import java.util.*;

public class Solution_78_Subsets {

    public static void main(String[] args) {
        Solution_78_Subsets solution_78_subsets = new Solution_78_Subsets();
        System.out.println(solution_78_subsets.subsets(new int[]{1, 2, 3}));//2^3
        System.out.println(solution_78_subsets.subsets(new int[]{1, 2, 3, 4}));//2^4
    }

    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> resultList = new ArrayList<>();
        resultList.add(new ArrayList<Integer>());
        for (int num : nums) {
            int size = resultList.size();//加上該元素之後組成新集合，同時還會與原集合結合，會有複製和引用的問題，此處用size非常巧妙
            for (int i = 0; i < size; i++) {
                List<Integer> subList = new ArrayList<>(resultList.get(i));
                subList.add(num);//該元素加入每個子集
                resultList.add(subList);//該新的子集加入新的結果集
            }
        }
        return resultList;
    }
}
 

輸出：

[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3], [4], [1, 4], [2, 4], [1, 2, 4], [3, 4], [1, 3, 4], [2, 3, 4], [1, 2, 3, 4]]