[LeetCode] 90.Subsets II tag: backtracking

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: [1,2,2]
Output:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

这个题目思路就是类似于DFS的backtracking. 是[LeetCode] 78. Subsets tag: backtracking 的update版本，或者说更加general的，因为允许有重复了。

所以在Subsets的基础/模板上，先sort nums，再加个判断，在for loop循环中，如果当前的跟前一个元素相同就跳过，相当于每次只取一次同样的数（在同一个for loop中）。

 

1. Constraints

1) edge case len(nums) == 0 => [ [ ] ]  但是还是可以

 

2. Ideas

DFS 的backtracking    T: O(2^n)     S; O(n)

 

3. Code 

 

1) using deepcopy, but the idea is obvious, use DFS [] -> [1] -> [1, 2] -> [1, 2, 3] then pop the last one, then keep trying until the end. 

import copy
class Solution:
    def subsets(self, nums: 'List [int]') -> 'List [ List [int] ]' :
        ans = []
        def helper(nums, ans, temp, pos):
            ans.append(copy.deepcopy(temp))
            for i in range(pos, len(nums)):
                if i > pos and nums[i] == nums[i - 1]:
                    continue
                temp.append(nums[i])
                helper(nums, ans, temp, i + 1)
                temp.pop()
        nums.sort()
        helper(nums, ans, [], 0)
        return ans

2) Use the same idea , but get rid of deepcopy

 

class Solution:
    def subsets(self, nums: 'List [int]') -> 'List [ List [int] ]' :
        ans = []
        def helper(nums, ans, temp, pos):
            ans.append(temp)
            for i in range(pos, len(nums)):
                if i > pos and nums[i] == nums[i - 1]:
                    continue
                helper(nums, ans, temp + [nums[i]], i + 1)
        nums.sort() # Optional
        helper(nums, ans, [], 0)
        return ans