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;    }};