Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

思路：采用深度优先搜索策略，先取1，然后从[2, 4]区间内取一个；取2，然后从[3, 4]区间取一个；取3，然后从[4]区间中取一个。

代码一：

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
        vector<vector<int>> result;
        if(n < k || k <= 0)   return result;
        
        return combine_helper(1, n, k);
    }
    
private:
    vector<vector<int>> combine_helper(int start, int end, int k)
    {
        vector<vector<int>> result;
        
        if(k == 1)
        {
            for(int i = start; i <= end; i++)
            {
                vector<int> temp;
                temp.push_back(i);
                result.push_back(temp);
            }
            return result;
        }
        
        for(int i = start; i <= end - k + 1; i++)
        {
            vector<vector<int>> temp;
            temp = combine_helper(i + 1, end, k - 1);
            for(int j = 0; j < temp.size(); j++)
            {
                temp[j].insert(temp[j].begin(), i);
                result.push_back(temp[j]);   
            }
        }
        
        return result;
    }
};

代码二：另一种相对比较简洁的递归写法

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
		vector<vector<int> > result;
		vector<int> path;
		dfs(n, k, 1, 0, path, result);
		return result;
    }
    
private:
	// start，开始的数, cur，已经选择的数目
	static void dfs(int n, int k, int start, int cur,
		vector<int> &path, vector<vector<int> > &result) 
	{
		if (cur == k) 
			result.push_back(path);

		for (int i = start; i <= n; ++i) 
		{
			path.push_back(i);
			dfs(n, k, i + 1, cur + 1, path, result);
			path.pop_back();
		}
	}
};