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subsets-ii.py 1.74 KB
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Allen Liu 提交于 2018-10-13 13:04 +08:00 . update
# Time: O(n * 2^n)
# Space: O(1)
class Solution(object):
def subsetsWithDup(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums.sort()
result = [[]]
previous_size = 0
for i in xrange(len(nums)):
size = len(result)
for j in xrange(size):
# Only union non-duplicate element or new union set.
if i == 0 or nums[i] != nums[i - 1] or j >= previous_size:
result.append(list(result[j]))
result[-1].append(nums[i])
previous_size = size
return result
# Time: O(n * 2^n) ~ O((n * 2^n)^2)
# Space: O(1)
class Solution2(object):
def subsetsWithDup(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
result = []
i, count = 0, 1 << len(nums)
nums.sort()
while i < count:
cur = []
for j in xrange(len(nums)):
if i & 1 << j:
cur.append(nums[j])
if cur not in result:
result.append(cur)
i += 1
return result
# Time: O(n * 2^n) ~ O((n * 2^n)^2)
# Space: O(1)
class Solution3(object):
def subsetsWithDup(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
result = []
self.subsetsWithDupRecu(result, [], sorted(nums))
return result
def subsetsWithDupRecu(self, result, cur, nums):
if not nums:
if cur not in result:
result.append(cur)
else:
self.subsetsWithDupRecu(result, cur, nums[1:])
self.subsetsWithDupRecu(result, cur + [nums[0]], nums[1:])
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LeetCode-Solutions
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