# Combination Sum III

### 描述​

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.

Ensure that numbers within the set are sorted in ascending order.

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]]

### 代码​

// Combination Sum III// Time Complexity: O(9*8*...*(10-k)), Space Complexity: O(k)public 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);        }    }}