Leetcode 78. Subsets

https://leetcode.com/problems/subsets/

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:

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

---

• 三种解法：回溯，迭代，位运算。搞清楚迭代和位运算的算法逻辑。
• 迭代就是在现有的结果集上扩展添加新元素，作为新的结果集。
• 位运算需要把所有结果集放在一起看，其实是判断二进制中第几位上为1，则该位所对应的数字在该结果集中。
• Python easy to understand solutions (DFS recursively, Bit Manipulation, Iteratively). - LeetCode Discuss
  • https://leetcode.com/problems/subsets/discuss/27301/Python-easy-to-understand-solutions-(DFS-recursively-Bit-Manipulation-Iteratively).
    
• C++ Recursive/Iterative/Bit-Manipulation - LeetCode Discuss
  • https://leetcode.com/problems/subsets/discuss/27278/C%2B%2B-RecursiveIterativeBit-Manipulation
• My solution using bit manipulation - LeetCode Discuss
  
  • https://leetcode.com/problems/subsets/discuss/27288/My-solution-using-bit-manipulation

class Solution:
    # Backtracking
    def subsets(self, nums: List[int]) -> List[List[int]]:
        if not nums:
            return None
        
        n_nums = len(nums)
        results = []
        
        def helper(index, path):
            results.append(path)
            
            for i in range(index, n_nums):
                helper( i + 1, path + [ nums[ i ] ] )
        
        helper(0, [])
        
        return results        

    # Iterative
    def subsets2(self, nums: List[int]) -> List[List[int]]:
        if not nums:
            return None
        
        results = [ [] ]
        
        for num in nums:
            results += [ item + [ num ] for item in results ]
        
        return results

    # Bit manipulation
    def subsets3(self, nums: List[int]) -> List[List[int]]:
        if not nums:
            return None
        
        results = []
        
        # check each subset
        for i in range( 1 << len( nums ) ):
            current = []

            # check if nums[j] is in current subset
            for j in range( len( nums) ):
                if (i >> j) & 1:
                    current.append( nums[ j ] )

            results.append(current)
        
        return results