leetcode array - power set

Power set. One way to write, all return a subset of the set. Collection does not contain duplicate elements.
Description: Solution Set can not contain duplicate subsets.

Example:

 输入： nums = [1,2,3]
 输出：
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

---

answer:

A thought: Binary Law

First know the number of the power set of a set of n elements as: n ^ 2
the nums = [l, 2,3], of length 3, for the nums [] position, the value Fu 1, Fu does not take 0.
Such as:
a first subset 0: 0 -> [0,0,0]: []
1 subset: 1 -> [0,0,1]: [nums [0]]
2 subsets: 2-> [0,1,0]: [nums [1 ]]
of three subsets: 3 -> [0, 1 , 1]: [nums [0], nums [1]]
4 subsets: 4 -> [ 1, 0, 0]: [ nums [2]]
5 subsets: 5 -> [1, 0 , 1]: [nums [0], nums [2]]
6 subsets: 6 -> [1 , 1, 0]: [nums [1], nums [2]]
7 subset: 7 -> [1, 1 , 1]: [nums [0], nums [1], nums [2]]

---

JAVA code:

class Solution {
        public List<List<Integer>> subsets(int[] nums) {
            // n个元素的集合，幂集个数为2^n
            
            List<List<Integer>> res = new ArrayList<>();
            int n = nums.length;
            //i代表第i个集合
            for (int i=0; i<(1<<n); i++){
                List<Integer> temp = new ArrayList<>();
				//j代表Nums[]的位置
                for(int j=0; j<n; j++){
                    if ((i & (1<<j))!=0) {
                        temp.add(nums[j]);
                    }
                }
                res.add(temp);
            }
            return res;
    }
}

---

Thinking two: enumeration method

= Nums [l, 2,3]

the JAVA code below

class Solution {
        public List<List<Integer>> subsets1(int[] nums) {
            List<List<Integer>> output = new ArrayList<>();
            //初始时output中只有一个空的list
            output.add(new ArrayList<>());

            for (int num: nums){
                List<List<Integer>> newSubsets = new ArrayList<>(); //存放每加入一个新元素后新增的子集

                //产生新的子集
                for (List<Integer> curr1: output) {
                    newSubsets.add(new ArrayList<Integer>(curr1){{add(num);}});  //这一句不是太懂
                }
                //将新子集加入到output中
                for (List<Integer> curr2: newSubsets)
                    output.add(curr2);
            }
            return output;
        }
    }

---

Thinking three: binomial tree

May constitute a subset of all full binary, and selected from the left and right subtrees considered problem is not selected, it indicates that each layer is selected and not selected for the i-th element. So as long as you can traverse the full binary tree.

JAVA code:

    public static void preOrder(List<List<Integer>> res, List<Integer> subset, int i, int[] nums){
        if (i>=nums.length) return;
        // 到了新的状态，记录新的路径，要重新拷贝一份
        subset = new ArrayList<Integer>(subset);

        res.add(subset);
        preOrder(res, subset, i+1, nums);
        subset.add(nums[i]);
        preOrder(res, subset, i+1, nums);
    }

Ideas Four: Backtracking

JAVA code:

    public static void backtrack(int[] nums, int i, List<Integer> sub, List<List<Integer>> res) {
        for (int j = i; j < nums.length; j++) {
            sub.add(nums[j]);
            res.add(new ArrayList<Integer>(sub));
            backtrack(nums, j + 1, sub, res);
            sub.remove(sub.size() - 1);
        }
    }