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]]

backtracking

time: C(m, k) = m! / k!(m-k)!

space: O(k + k * # of ans)

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> res = new ArrayList<>();
        backtracking(k, n, 1, new ArrayList<>(), res);
        return res;
    }
    
    private void backtracking(int k, int n, int start, List<Integer> tmp, List<List<Integer>> res) {
        if(k == 0) {
            if(n == 0)
                res.add(new ArrayList<>(tmp));
        }
        
        for(int i = start; i <= 9; i++) {
            if(i > n) break;
            tmp.add(i);
            backtracking(k - 1, n - i, i + 1, tmp, res);
            tmp.remove(tmp.size() - 1);
        }
    }
}

二刷：

time: O(2 ^ k), space: O(k)

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> res = new ArrayList<>();
        dfs(k, n, 1, new ArrayList<>(), res);
        return res;
    }
    
    public void dfs(int k, int target, int start, List<Integer> list, List<List<Integer>> res) {
        if(k == 0) {
            if(target == 0) {
                res.add(new ArrayList<>(list));
            }
            return;
        }
        
        for(int i = start; i <= 9; i++) {
            if(i > target) {
                break;
            }
            list.add(i);
            dfs(k - 1, target - i, i + 1, list, res);
            list.remove(list.size() - 1);
        }
    }
}