## [Leetcode]216. Combination Sum III(C++)

### 题目描述

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.

### 例子

#### 例子 1

Input: `k = 3, n = 7` Output: `[[1,2,4]]` Explanation: 1 + 2 + 4 = 7 There are no other valid combinations.

#### 例子 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.

#### 例子 3

Input: `k = 4, n = 1` Output: `[]` Explanation: There are no valid combinations. `[1,2,1]` is not valid because 1 is used twice.

### Constraints

• `2 <= k <= 9`
• `1 <= n <= 60`

### 解题思路

``````#include <vector>

class Solution {
public:
std::vector<std::vector<int>> combinationSum3(int k, int n) {
k_ = k;
std::vector<int> comb;
std::vector<std::vector<int>> result;
dfs(comb, 1, n, result);
return result;
}

private:
int k_;

void dfs(std::vector<int>& comb, int idx, int target,
std::vector<std::vector<int>>& result) {
if (comb.size() == k_) {
if (target == 0) {
result.push_back(comb);
}
return;
}
if (target < 0) {
return;
}
for (int i = idx; i <= 9; i++) {
comb.push_back(i);
dfs(comb, i + 1, target - i, result);
comb.pop_back();
}
}
};
``````
• 时间复杂度: O(n!)
• 空间复杂度: O(n!)

GitHub 代码同步地址： 216.CombinationSumIii.cpp

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