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], ]
Analyse:
Backtracking. For each number i from 1...n, choose i and fix it, for i + 1...n, do the same thing until the temp vector's size equals to k. Be cautious that we can't use depth == k as the condition, because the depth is always changing and the amount of numbers we want to pick up is no larger than n. So the depth information is not accurate. Suppose k = 3, n = 8, and we encounter 1, 2, 8 as the temp vector, the depth is 8 while it doesn't equals to k. However, we can intuitively realize that this vector should be placed into the result.
1 class Solution { 2 public: 3 vector<vector<int>> combine(int n, int k) { 4 vector<vector<int> > result; 5 if(n < k || n <= 0) return result; 6 7 vector<int> temp; 8 helper(result, temp, n, k, 1); 9 return result; 10 } 11 12 void helper(vector<vector<int> >& result, vector<int>& temp, int n, int k, int depth){ 13 if(temp.size() == k){ 14 result.push_back(temp); 15 return; 16 } 17 18 for(int i = depth; i <= n; i++){ 19 temp.push_back(i); 20 helper(result, temp, n, k, i + 1); 21 temp.pop_back(); 22 } 23 } 24 };