题目描述

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

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

Example
If S = [1,2,3], a solution is:

[
 [3],
 [1],
 [2],
 [1,2,3],
 [1,3],
 [2,3],
 [1,2],
 []
]
Challenge
Can you do it in both recursively and iteratively?

大致是给一个无序数组，让你求它的子集

递归解法

先说说递归吧，比较简单
递归出口

        if (nums.length == n) {
            result.add(new ArrayList<>(ints));
            return;
        }

将数组集合中的每个元素依次加进去，再拿出来继续进行递归

		ints.add(nums[n]);
        helper(nums, ints, result, n + 1);
        ints.remove(ints.size() - 1);
        helper(nums, ints, result, n + 1);

全部代码如下

public List<List<Integer>> subsets(int[] nums) {
        // write your code here
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums);
        helper(nums, new ArrayList<>(), result, 0);
        return result;
    }

    private void helper(int[] nums,
                        ArrayList<Integer> ints,
                        List<List<Integer>> result,
                        int n) {
        //递归出口
        if (nums.length == n) {
            result.add(new ArrayList<>(ints));
            return;
        }
        ints.add(nums[n]);
        helper(nums, ints, result, n + 1);
        ints.remove(ints.size() - 1);
        helper(nums, ints, result, n + 1);
    }

非递归解法

非递归解法是这样的，用二进制01代表该数组元素有没有被选中，然后判断该元素需不需要添加进去
举例，nums[1,2]
结果如下：

代码如下

public List<List<Integer>> subsets(int[] nums) {
        // write your code here
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums);
        for (int i = 0; i < (1 << nums.length); i++) {
            List<Integer> n = new ArrayList<>();
            for (int j = 0; j < nums.length; j++) {
                if ((i & (1 << j)) != 0) {
                    n.add(nums[j]);
                }
            }
        }
        return result;
    }