Combination Sum III

题目：

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.

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]]

分析：

这道题目来自于leetcode。有时间了要分析一下怎么用const number来表示。

解法：

class Solution {
public:

    int getSum(vector<int> instance) {
        int sum = 0;
        for (int i = 0; i < instance.size(); i++) {
            sum += instance[i];
        }
        return sum;
    }

    void dfs(vector<int> instance, vector<vector<int> >& res, int start, int n, int k) {

        if (instance.size() == k) {
            int sum = getSum(instance);
            if (sum > n) {
                return;
            } else if (sum == n) {
                res.push_back(instance);
                return;
            }
        }

        for (int i = start; i <= 9; i++) {
            instance.push_back(i);
            dfs(instance, res, i + 1, n, k);
            instance.pop_back();
        }
    }

public:
    vector<vector<int>> combinationSum3(int k, int n) {
        if (k < 1 || k > 9) {
            return vector<vector<int> >();
        }

        if (n < 1 || n > 45) {
            return vector<vector<int> >();
        }

        vector<vector<int> > res;
        vector<int> instance;

        dfs(instance, res, 1, n, k);

        return res;
    }
};