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Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note:
For example,
If nums = [1,2,2], a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
解题思路一：
偷懒做法，将Java for LeetCode 078 Subsets中的List换为Set即可通过测试，JAVA实现如下：
public List<List<Integer>> subsetsWithDup(int[] nums) {
Set<List<Integer>> list = new HashSet<List<Integer>>();
list.add(new ArrayList<Integer>());
Arrays.sort(nums);
for(int i=1;i<=nums.length;i++)
dfs(list, nums.length, i, 0,nums,-1);
return new ArrayList(list);
}
static List<Integer> alist = new ArrayList<Integer>();
static void dfs(Set<List<Integer>> list, int n, int k, int depth,int[] nums,int last) {
if (depth >= k) {
list.add(new ArrayList<Integer>(alist));
return;
}
for (int i = last+1; i <= n-k+depth; i++) {
alist.add(nums[i]);
dfs(list, n, k, depth + 1,nums,i);
alist.remove(alist.size() - 1);
}
}
Java for LeetCode 090 Subsets II
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原文地址：http://www.cnblogs.com/tonyluis/p/4516099.html
