Leetcode Subsets II

Leetcode Subsets II，本问题与Subsets 类似，只是需要处理重复子集问题，这个问题，我们可以使用每次从长度为m的子集构成长度为m+1的子集时，只对等值的元素添加一次，这样就可以保证没有重复现象，处理代码与Subsets类似，如下：

#include<iostream>#include<vector>#include<algorithm>using namespace std;class Solution {public:    vector<vector<int> > subsetsWithDup(vector<int>& nums) {        // Sort the nums for the left solution        sort(nums.begin(), nums.end());        int len = nums.size();        // Init the result vector re        vector<vector<int> > re;        re.push_back(vector<int>());        // The pos vector saves the last element postion of the re[i]        // in the nums.        vector<int> pos;        // The first element in re is empty, so the first value is -1        pos.push_back(-1);        // The start and end position of the length k element in re        int start = 0;        int end = 0;        while (re.back().size() != len) {            // The elements of re with length k can be formed by the elements            // of re with length k-1 append an element after its last element            for (int i = start; i <= end; i ++) {                for (int j = pos[i] + 1; j < len; j ++) {                    // Every time construct the element with length k                    // for element with length k-1 only include the elements                    // in nums with equaling values once                    if ((j == pos[i] + 1) || (j != 0 && nums[j] != nums[j - 1])) {                        vector<int> temp(re[i]);                        temp.push_back(nums[j]);                        re.push_back(temp);                        pos.push_back(j);                    }                }            }            //Update the start and end position of elements with length k            start = end + 1;            end = re.size() - 1;        }        return re;    }};int main(int argc, char* argv[]) {    Solution so;    vector<int> test;    test.push_back(2);    test.push_back(1);    test.push_back(2);    test.push_back(3);    vector<vector<int> > re = so.subsetsWithDup(test);    for (int i = 0; i < re.size(); i ++) {        for (int j = 0; j < re[i].size();j ++) {            cout<<re[i][j]<<" ";        }        cout<<endl;    }    return 0;}