【Subsets】cpp

题目：

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

Note:

• Elements in a subset must be in non-descending order.
• 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],
  []
]

代码：

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
            const int len = nums.size();
            std::sort(nums.begin(), nums.end());
            vector<vector<int> > ret = Solution::subsubsets(nums, 0, len-1);
            vector<int> none;
            ret.push_back(none);
            return ret;
    }
    static vector<vector<int> > subsubsets(vector<int>& nums, int begin, int end)
    {
            vector<vector<int> > ret;
            if ( begin>end ) return ret;
            for ( int i = begin; i <=end; ++i )
            {
                // puts the curr value in
                vector<int> curr;
                curr.push_back(nums[i]);
                ret.push_back(curr);
                // get the subset
                vector<vector<int> > subset = Solution::subsubsets(nums, i+1, end);
                for ( int j = 0; j < subset.size(); ++j )
                {
                    //ret.push_back(subset[j]);
                    subset[j].insert(subset[j].begin(), nums[i]);
                    ret.push_back(subset[j]);
                }
            }
            return ret;
    }
};

tips：

每层递归的任务是活的传入数组的全部子集

1. 留出第i个元素

2. 把i+1到end的元素送到下一层递归

3. 递归终止条件是begin>end

返回下一集全部子集后，算上第i个元素后的全部集合如下：

1）第i个元素单独算一个

2）第i个元素 + 子集中每个元素

这样递归到第0个元素，就得到了全部的非空子集；再根据题目要求补上空集。

============================================

完成之后觉得好像哪里不对。{1，2，3} 如果留下了1传入{2，3}时，这个过程中已经获得了{2，3}全部子集。

修改后的代码如下：

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
            const int len = nums.size();
            std::sort(nums.begin(), nums.end());
            vector<vector<int> > ret = Solution::subsubsets(nums, 0, len-1);
            vector<int> none;
            ret.push_back(none);
            return ret;
    }
    static vector<vector<int> > subsubsets(vector<int>& nums, int begin, int end)
    {
            vector<vector<int> > ret;
            if ( begin>end ) return ret;
            // puts the curr value in
            vector<int> curr;
            curr.push_back(nums[begin]);
            ret.push_back(curr);
            // get the subset
            vector<vector<int> > subset = Solution::subsubsets(nums, begin+1, end);
            for ( int j = 0; j < subset.size(); ++j )
            {
                ret.push_back(subset[j]);
                subset[j].insert(subset[j].begin(), nums[begin]);
                ret.push_back(subset[j]);
            }
            return ret;
    }
};

tips：

这个版本整理了逻辑误区。