Combination Sum III

挺简单的一道题，也出了3个错误，如何bug free啊

public class Solution {    public List<List<Integer>> combinationSum3(int k, int n) {        List<List<Integer>> result = new LinkedList<>();        if (k == 0 || n < 1) {            return result;        }        List<Integer> list = new LinkedList<>();        //1 combinationSum3Helper(result, list, k, n, 0);        combinationSum3Helper(result, list, k, n, 1);        return result;    }        private void combinationSum3Helper(List<List<Integer>> result, List<Integer> list, int k, int n, int pos) {        if (k == 0 && n == 0) {            result.add(new LinkedList<>(list));            return;        } else if (n < 0 || k < 0) {            return;        }        //2 for (int i = pos; i <= n; i++) {        for (int i = pos; i <= 9; i++) {            list.add(i);            //3 ombinationSum3Helper(result, list, k - 1, n - i, i++);            combinationSum3Helper(result, list, k - 1, n - i, i + 1);            list.remove(list.size() - 1);        }            }}

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.

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]]