Leetcode Subsets

Given a set of distinct integers,nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

[  [3],  [1],  [2],  [1,2,3],  [1,3],  [2,3],  [1,2],  []]

第一个想到的就是递归大法，代码如下：

class Solution {public:    vector<vector<int>> subsets(vector<int>& nums) {        vector<vector<int>> result;        vector<int> temp;        sub(nums,result,temp,0);        return result;    }        void sub(vector<int>& nums,vector<vector<int>>& result,vector<int>& temp,int start)    {        result.push_back(temp);        for(int i=start;i<nums.size();i++)        {            temp.push_back(nums[i]);            sub(nums,result,temp,i+1);            temp.pop_back();        }    }};

递归必然可以转换为迭代，一般来说，迭代的开销比递归中的函数调用成本小，所以迭代更有效。迭代一般要通过栈结构，但本题的迭代特别有意思，迭代代码如下：

class Solution {public:    vector<vector<int>> subsets(vector<int>& nums) {        vector<vector<int>> res(1, vector<int>());        int n = nums.size();        if (n == 0) return res;        sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size(); i++) {            int n = res.size();            for (int j = 0; j < n; j++) {                res.push_back(res[j]);                res.back().push_back(nums[i]);            }        }        return res;    }};