216. Combination Sum III

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Note:

• All numbers will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        final List<List<Integer>> result = new ArrayList<>();
        final List<Integer> path = new ArrayList<>();
        dfs(k, n, path, result);
        return result;
    }
    private static void dfs(int step, int gap, List<Integer> path,
                            List<List<Integer>> result) {
        if (step == 0) {
            if (gap == 0) {
                result.add(new ArrayList<>(path));
            }
            return;
        }

        if (gap < 1) return;

        final int start = path.isEmpty() ? 1 : path.get(path.size() - 1)+1;
        for (int i = start; i < 10; ++i) {
            path.add(i);
            dfs(step - 1, gap - i, path, result);
            path.remove(path.size() - 1);
        }
    }
}

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        final List<List<Integer>> result = new ArrayList<>();
        final List<Integer> path = new ArrayList<>();
        dfs(result, path, k, 1, n);
        return result;
    }
    private static void dfs(List<List<Integer>> ans, List<Integer> comb, int k,  int start, int n) {
        if (comb.size() > k) {
            return;
        }
        if (comb.size() == k && n == 0) {
            List<Integer> li = new ArrayList<Integer>(comb);
            ans.add(li);
            return;
        }
        for (int i = start; i <= n && i<=9; i++) {
                
                comb.add(i);
                dfs(ans, comb, k, i+1, n-i);
                comb.remove(comb.size() - 1);
            
        }
    }
}