90 Subsets II 子集 II
Description:
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],
[]
]
题目描述:
给定一个可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。
说明：
解集不能包含重复的子集。
示例 :
输入: [1,2,2]
输出:
[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]
思路:
参考LeetCode #78 Subsets 子集和LeetCode #47 Permutations II 全排列 II
在回溯的基础上加上去重
时间复杂度O(n * 2 ^ n), 空间复杂度O(n * 2 ^ n)
代码:
C++:
class Solution
{
public:
vector> subsetsWithDup(vector& nums)
{
sort(nums.begin(), nums.end());
int n = nums.size();
vector > result;
for (int i = 0; i < 1 << n; ++i)
{
vector temp;
bool flag = true;
for (int j = 0; j < n; j++)
{
if (i & (1 << j))
{
if (j != 0 and nums[j] == nums[j - 1] and ((i & (1 << (j - 1))) == 0))
{
flag = false;
break;
}
temp.push_back(nums[j]);
}
}
if (flag) result.push_back(temp);
}
return result;
}
};
Java:
class Solution {
public List> subsetsWithDup(int[] nums) {
List> result = new ArrayList<>((int)Math.pow(2, nums.length));
List temp = new ArrayList<>(nums.length);
Arrays.sort(nums);
backtrack(result, temp, nums, 0);
return result;
}
private void backtrack(List> result, List temp, int[] nums, int start) {
result.add(new ArrayList<>(temp));
for (int i = start; i < nums.length; i++) {
if (i != start && nums[i] == nums[i - 1]) continue;
temp.add(nums[i]);
backtrack(result, temp, nums, i + 1);
temp.remove(temp.size() - 1);
}
}
}
Python:
class Solution:
def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
result, temp = [], []
nums.sort()
def backtrack(start: int) -> None:
result.append(temp[:])
for i in range(start, len(nums)):
if i != start and nums[i] == nums[i - 1]:
continue
temp.append(nums[i])
backtrack(i + 1)
temp.pop()
backtrack(0)
return result