Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
- Elements in a subset must be in non-descending order.
- The solution set must not contain duplicate subsets.
For example,
If S = [1,2,2], a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
Solution 1:
class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int> &S) {
sort(S.begin(), S.end());
vector<vector<int>> result;
int len = S.size();
int maxLen = pow(2, len);
for(int i = 0; i < maxLen; i++)
{
vector<int> temp;
for(int j = 0; j < len; j++)
{
if(i & (1 << j))
temp.push_back(S[j]);
}
if(find(result.begin(), result.end(), temp) == result.end())
result.push_back(temp);
}
return result;
}
};Solution 2:
class Solution {
public:
vector<vector<int>> result;
void generate(vector<int> &S, vector<int> temp, int level)
{
if(level == S.size())
{
if(count(result.begin(), result.end(), temp) == 0)
result.push_back(temp);
return;
}
generate(S, temp, level+1);
temp.push_back(S[level]);
generate(S, temp, level+1);
}
vector<vector<int> > subsetsWithDup(vector<int> &S) {
sort(S.begin(), S.end());
result.clear();
vector<int> temp;
generate(S, temp, 0);
return result;
}
};Solution 3:
class Solution {
public:
set<vector<int>> result;
void generate(vector<int> &S, vector<int> temp, int level)
{
if(level == S.size())
{
result.insert(temp);
return;
}
generate(S, temp, level+1);
temp.push_back(S[level]);
generate(S, temp, level+1);
}
vector<vector<int> > subsetsWithDup(vector<int> &S) {
sort(S.begin(), S.end());
result.clear();
vector<int> temp;
generate(S, temp, 0);
vector<vector<int>> ret(result.begin(), result.end());
return ret;
}
};

本文介绍了一种处理包含重复元素的集合并返回所有可能子集的算法。通过三种不同的实现方式,确保子集中元素按非递减顺序排列且不包含重复的子集。示例中使用了[1,2,2]作为输入数据,展示了预期的输出结果。
499

被折叠的 条评论
为什么被折叠?



