LeetCode 216. Combination Sum III

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本文介绍了一个算法问题：寻找所有可能的k个数的组合，这些数加起来等于给定的数n，且只允许使用1到9之间的数字。每个组合中的数字必须唯一，并按升序排列。文章提供了具体的实现代码及示例。

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]]

I first allocate an array to store [1 ..9]. Actually this is not needed...

#include <vector>
#include <iostream>
using namespace std;

// all possible k numbers that add up to a number n, given that only numbers from 1 to 9.
// Input: k = 3, n = 7
// output: [[1, 2, 4]]

void combinationSum3(int& k, int& val, int index, int target, vector<int>& path, vector< vector<int> >& res) {
    if(k < 0) return;
    if(val == target && k == 0) {
        res.push_back(path);
    }
    for(int i = index; i < 10; ++i) {
        val += i;
        k--;
        path.push_back(i);
        combinationSum3(k, val, i + 1, target, path, res);
        val -= i;
        k++;
        path.pop_back();
    }
}

vector< vector<int> > combinationSum3(int k, int n) {
    vector< vector<int> > res;
    vector<int> path;
    int sum = 0;
    combinationSum3(k, sum, 1, n, path, res);
    return res;
}

int main(void) {
    vector< vector<int> > res = combinationSum3(3, 7);
    for(int i = 0; i < res.size(); ++i) {
        for(int j = 0; j < res[0].size(); ++j) {
            cout << res[i][j] << endl;
        }
        cout << endl;
    }
}