Leetcode Subsets

Leetcode Subsets 相关解决方案，本方法使用迭代完成算法，在leetcode上处理时间为4ms，以下提供相关代码以及测试：

#include<iostream>#include<vector>#include<algorithm>using namespace std;class Solution {public:    vector<vector<int> > subsets(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 ++) {                    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(1);    test.push_back(2);    test.push_back(3);    vector<vector<int> > re = so.subsets(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;}