LeetCode之Subsets II

//递归解法
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
vector<vector<int> > res;
if(nums.empty()) return res;
sort(nums.begin(), nums.end());

for(int count = 0; count < nums.size()+1; ++count){
vector<int> tmp;
subsets_sub(nums, res, tmp, 0, count);
}
return res;
}

void subsets_sub(const vector<int> &nums, vector<vector<int> > &res,
vector<int> &tmp_res, int start, int count){
/*函数功能：递归获取包含count个数的子数组。*/
if(count == 0){
res.push_back(tmp_res);
return;
}
for(int i = start; i < nums.size();){
if(count <= nums.size()-i){
int n(1);
for(int j = i+1; j < nums.size() && nums[j] == nums[i]; ++j) ++n;

for(int j = 1; j <= n; ++j){
int k;
for(k = 1; k <= j; ++k) tmp_res.push_back(nums[i]);
subsets_sub(nums, res, tmp_res, i+n, count-j);
while(--k > 0) tmp_res.pop_back();
}
i += n;
}
else break;
}
}
};

//迭代求解
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
vector<vector<int> > res;
if(nums.empty()) return res;
sort(nums.begin(), nums.end());

res.push_back(vector<int>());
for(int i = 0; i < nums.size();){
int n(0);
for(int j = i; j < nums.size() && nums[j] == nums[i]; ++j) ++n;

int size = res.size();
for(int k = 1; k < n+1; ++k){
res.reserve(res.size()+size);
for(int j = 0; j < size; ++j){
res.push_back(res[(k-1)*size + j]);
res[k*size + j].push_back(nums[i]);
}
}
i += n;
}
return res;
}
};