README_EN.md

216. Combination Sum III

中文文档

Description

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. [1,2,1] is not valid because 1 is used twice.

Example 4:

Input: k = 3, n = 2
Output: []
Explanation: There are no valid combinations.

Example 5:

Input: k = 9, n = 45
Output: [[1,2,3,4,5,6,7,8,9]]
Explanation:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
​​​​​​​There are no other valid combinations.

 

Constraints:

• 2 <= k <= 9
• 1 <= n <= 60

Solutions

DFS.

Python3

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        def dfs(i, s, t):
            if i > 9 or s > n or len(t) > k:
                return
            if s == n and len(t) == k:
                ans.append(t[:])
                return
            i += 1
            t.append(i)
            dfs(i, s + i, t)
            t.pop()
            dfs(i, s, t)

        ans = []
        dfs(0, 0, [])
        return ans

Java

class Solution {
    private List<List<Integer>> ans;

    public List<List<Integer>> combinationSum3(int k, int n) {
        ans = new ArrayList<>();
        dfs(0, n, k, new ArrayList<>());
        return ans;
    }

    private void dfs(int i, int n, int k, List<Integer> t) {
        if (i > 9 || n < 0 || t.size() > k) {
            return;
        }
        if (n == 0 && t.size() == k) {
            ans.add(new ArrayList<>(t));
            return;
        }
        ++i;
        t.add(i);
        dfs(i, n - i, k, t);
        t.remove(t.size() - 1);
        dfs(i, n, k, t);
    }
}

C++

class Solution {
public:
    vector<vector<int>> ans;

    vector<vector<int>> combinationSum3(int k, int n) {
        vector<int> t;
        dfs(0, n, k, t);
        return ans;
    }

    void dfs(int i, int n, int k, vector<int>& t) {
        if (i > 9 || n < 0 || t.size() > k) return;
        if (n == 0 && t.size() == k)
        {
            ans.push_back(t);
            return;
        }
        ++i;
        t.push_back(i);
        dfs(i, n - i, k, t);
        t.pop_back();
        dfs(i, n, k, t);
    }
};

Go

func combinationSum3(k int, n int) [][]int {
	var ans [][]int
	var t []int
	var dfs func(i, n int, t []int)
	dfs = func(i, n int, t []int) {
		if i > 9 || n < 0 || len(t) > k {
			return
		}
		if n == 0 && len(t) == k {
			cp := make([]int, len(t))
			copy(cp, t)
			ans = append(ans, cp)
			return
		}
		i++
		t = append(t, i)
		dfs(i, n-i, t)
		t = t[:len(t)-1]
		dfs(i, n, t)
	}

	dfs(0, n, t)
	return ans
}

...