<p>Given two integers <code>n</code> and <code>k</code>, return <em>all possible combinations of</em> <code>k</code> <em>numbers chosen from the range</em> <code>[1, n]</code>.</p>

<p>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, k = 2
<strong>Output:</strong> [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
<strong>Explanation:</strong> There are 4 choose 2 = 6 total combinations.
Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1, k = 1
<strong>Output:</strong> [[1]]
<strong>Explanation:</strong> There is 1 choose 1 = 1 total combination.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 20</code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
</ul>

class Solution {
public:
    void func(vector<int>& num, int k, int i, vector<vector<int>>& ans,
              vector<int>& tp) {
        if (tp.size() == k) {
            ans.push_back(tp);
            return;
        }
        if(num.size()-i+tp.size()<k){
            return;
        }
        if(num.size()-i+tp.size()==k){
            vector<int> temp=tp;
            for(int j=i;j<num.size();j++) temp.push_back(num[j]);
            ans.push_back(temp);
            return;
        }
        tp.push_back(num[i]);
        func(num,k,i+1,ans,tp);
        tp.pop_back();
        func(num,k,i+1,ans,tp);
    }
    vector<vector<int>> combine(int n, int k) {
        vector<vector<int>> ans;
        vector<int> temp;
        vector<int> num(n);
        for(int i=0;i<n;i++) num[i]=i+1;
        func(num,k,0,ans,temp);
        return ans;
    }
};
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