Difficulty: Medium
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],
[]
]
Language: Java
class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> elem = new ArrayList<>();
result.add(elem);
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
int dupCount = 0;
while (((i + 1) < nums.length) && nums[i + 1] == nums[i]) {
dupCount++;
i++;
}
int size = result.size();
for (int j = 0; j < size; j++) {
elem = new ArrayList<>(result.get(j));
for (int k = 0; k <= dupCount; k++) {
elem.add(nums[i]);
result.add(new ArrayList<>(elem)); // 注意这里是要放在 for 循环里面的
}
}
}
return result;
}
}