216. Combination Sum III
     Find all valid combinations of k numbers that sum up to n such that the following conditions are true:
     Only numbers 1 through 9 are used.
     Each number is used at most once.
     Return a list of all possible valid combinations.
     The list must not contain the same combination twice, and the combinations may be returned in any order.

Example 1:

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

Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.

Example 2:

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

Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.

Example 3:

Input: k = 4, n = 1
Output: []

Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Constraints:
2 <= k <= 9
1 <= n <= 60

给 元素重复 无序 数组，求满足条件的子数组，组合问题
需要子数组满足的条件：子数组元素和等于某个数

package leetcode.problems.medium;

import java.util.ArrayList;
import java.util.List;

public class CombinationSumIII {
    public List<List<Integer>> combinationSum3(int k, int n) {
        int[] nums = {1,2,3,4,5,6,7,8,9};
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        backtrack(result, path, nums, 0, k, n);
        return result;
    }
    private void backtrack(List<List<Integer>> result, List<Integer> path, int[] nums, int start, int level, int stillNeed) {
        if (stillNeed == 0 && level == 0) {
            result.add(new ArrayList<>(path));
            return;
        } else if (level < 0 || stillNeed <= 0) {
            return;
        }
        for (int i = start; i < nums.length; i++) {
            path.add(nums[i]);
            backtrack(result, path, nums, i+1, level-1, stillNeed-nums[i]);
            path.remove(path.size()-1);
        }
    }
}