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); } } }