90-Subsets II

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题意:

找比当前组合大的那个最小的组合

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], [] ]

思路

• 回溯法
• 备选元素有重复：用used boolean数组控制
• 结果中不能有重复：不走重复路线，用index控制

class Solution(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []      
        if not nums: return res
        nums.sort()
        
        def backtrack(res, temp, nums, used, index):
            res.append(temp[:])
            
            for i in range(index, len(nums)): # 此处易错，务必用上index
                if used[i]: continue
                if (i > 0 and nums[i] == nums[i-1] and not used[i-1]): continue
                temp.append(nums[i])
                used[i] = True
                backtrack(res, temp, nums, used, i+1)
                used[i] = False
                temp.pop()
                
        used = [False] * len(nums)
        backtrack(res, [], nums, used, 0)
        return res

分析

• Time: O（\(2^n\)）
• Space：O(n)