[LeetCode] 90.Subsets II tag: backtracking
2019-04-11 10:21 Johnson_强生仔仔 阅读(259) 评论(0) 编辑 收藏 举报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)。
Note: for input [4,4,4,1,4], the output is [[],[1],[1,4],[1,4,4],[1,4,4,4],[1,4,4,4,4],[4],[4,4],[4,4,4],[4,4,4,4]]
if we use only visited = set() before the for loop in the helper function without nums.sort(), we will get results like below:
[[],[4],[4,4],[4,4,4],[4,4,4,1],[4,4,4,1,4],[4,4,4,4],[4,4,1],[4,4,1,4],[4,1],[4,1,4],[1],[1,4]]
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() helper(nums, ans, [], 0) return ans
3) use visited set to get only one from the duplicates
class Solution: def subsetsWithDup(self, nums: List[int]) -> List[List[int]]: ans = [] def helper(nums, ans, temp, pos): ans.append(temp) visited = set() for i in range(pos, len(nums)): if nums[i] not in visited: helper(nums, ans, temp + [nums[i]], i + 1) visited.add(nums[i]) nums.sort() helper(nums, ans, [], 0) return ans