Description
Given an integer array nums that may contain duplicates, return all possible subsets (the power set).
The solution set must not contain duplicate subsets. Return the solution in any order.
Example 1:
Input: nums = [1,2,2] Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]
Example 2:
Input: nums = [0] Output: [[],[0]]
Constraints:
1 <= nums.length <= 10-10 <= nums[i] <= 10
Solution
Python3
class Solution:
def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
n = len(nums)
nums.sort()
ans = [[]]
start = 1
for i in range(n):
temp = []
for j, num in enumerate(ans):
if i > 0 and nums[i] == nums[i-1] and j < start: continue
c = num + [nums[i]]
temp.append(c)
start = len(ans)
ans += temp
return ans
C++
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
vector<vector<int>> res;
for (int i=0; i < 1<<n; ++i){
vector<int> c;
bool illegal = false;
for (int j = 0; j < n; ++j)
if (i >> j&1){
if (j > 0 && nums[j]==nums[j-1] && (i>>(j-1)&1) == 0){
illegal = true;
break;
}else{
c.push_back(nums[j]);
}
}
if (!illegal){
res.push_back(c);
}
}
return res;
}
};Python
class Solution:
def subsetsWithDup(self, nums):
if not nums:
return []
nums.sort()
res, cur = [[]], []
for i in range(len(nums)):
if i > 0 and nums[i] == nums[i-1]:
cur = [item + [nums[i]] for item in cur]
else:
cur = [item + [nums[i]] for item in res]
res += cur
return res