permutations-ii.py

# Time:  O(n * n!)
# Space: O(n)
#
# Given a collection of numbers that might contain duplicates, return all possible unique permutations.
# 
# For example,
# [1,1,2] have the following unique permutations:
# [1,1,2], [1,2,1], and [2,1,1].
#

class Solution(object):
    def permuteUnique(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        result = []
        used = [False] * len(nums)
        self.permuteUniqueRecu(result, used, [], nums)
        return result
    
    def permuteUniqueRecu(self, result, used, cur, nums):
        if len(cur) == len(nums):
            result.append(cur + [])
            return
        for i in xrange(len(nums)):
            if used[i] or (i > 0 and nums[i-1] == nums[i] and not used[i-1]):
                continue
            used[i] = True
            cur.append(nums[i])
            self.permuteUniqueRecu(result, used, cur, nums)
            cur.pop()
            used[i] = False
                
class Solution2:
    # @param num, a list of integer
    # @return a list of lists of integers
    def permuteUnique(self, nums):
        solutions = [[]]
        
        for num in nums:
            next = []
            for solution in solutions:
                for i in xrange(len(solution) + 1):
                    candidate = solution[:i] + [num] + solution[i:]
                    if candidate not in next:
                        next.append(candidate)
                
            solutions = next 
            
        return solutions

if __name__ == "__main__":
    print Solution().permuteUnique([1, 1, 2])
    print Solution().permuteUnique([1, -1, 1, 2, -1, 2, 2, -1])