[LeetCode] Combination Sum III 組合之和之三

 

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]]

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

 

這道題題是組合之和系列的第三道題，跟之前兩道 Combination Sum，Combination Sum II 都不太一樣，那兩道的聯系比較緊密，變化不大，而這道跟它們最顯著的不同就是這道題的個數是固定的，為k。個人認為這道題跟那道 Combinations 更相似一些，但是那道題只是排序，對k個數字之和又沒有要求。所以實際上這道題是它們的綜合體，兩者雜糅到一起就是這道題的解法了，n是k個數字之和，如果n小於0，則直接返回，如果n正好等於0，而且此時out中數字的個數正好為k，說明此時是一個正確解，將其存入結果res中，具體實現參見代碼入下：

 

class Solution {
public:
    vector<vector<int> > combinationSum3(int k, int n) {
        vector<vector<int> > res;
        vector<int> out;
        combinationSum3DFS(k, n, 1, out, res);
        return res;
    }
    void combinationSum3DFS(int k, int n, int level, vector<int> &out, vector<vector<int> > &res) {
        if (n < 0) return;
        if (n == 0 && out.size() == k) res.push_back(out);
        for (int i = level; i <= 9; ++i) {
            out.push_back(i);
            combinationSum3DFS(k, n - i, i + 1, out, res);
            out.pop_back();
        }
    }
};

 

類似題目：

Combination Sum IV 

Combination Sum II

Combination Sum

 

LeetCode All in One 題目講解匯總(持續更新中...)