LeetCode Problem 90. Subsets II

python solution

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class (object): def subsetsWithDup(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ """ 思路整理:DFS recursion 1)对nums进行排序以避免nums中重复元素不聚集在一起如[1,2,4,4,3,4] 2)从nums中依次取出元素加到path中进行DFS 3)对于重复元素如[1,2,2]中的2，res中每个以2为开头的数组的只取nums的第一个2 Explanation 1)sort nums to gather all same number together 2)take numbers from nums iteratively and use them to conduct DFS 3)for same numbers like 2 in [1,2,2],each array only take the first 2 as its first 2 """ def helper(start,end,path,res): res.append(path) for i in range(start,end): if i!=start and nums[i]==nums[i-1]:continue helper(i+1,end,path+[nums[i]],res) return res=[] nums.sort() helper(0,len(nums),[],res) return res< 大专栏  LeetCode Problem 90. Subsets IIbr/>

java solution

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对于无重复元素的n维数组 有2^n个子数组 因为每个元素在每个子数组里都只有出现或者不出现两种可能 而对于有重复元素的数组如[1,2,2] (1)[] (2)[] [1] 复制一份[] 并向其中加入1 (3)[] [1] [2] [1,2] 复制一份[] [1] 并向其中加入2 (3)[] [1] [2] [1,2] [2,2] [1,2,2]对于重复元素 只向前一次添加过与其相同元素的数组中添加该元素 复制一份[2] [1,2] 并向其中加入2 */ class { public List> subsetsWithDup(int[] nums) { List> res=new ArrayList>(); int start=0; res.add(new ArrayList()); Arrays.sort(nums); for(int i=0;i { if(i==0||nums[i]!=nums[i-1]) start=0; int size=res.size(); for(int j=start;j { List temp=new ArrayList(res.get(j)); temp.add(nums[i]); res.add(temp); } start=size; } return res; } }