216. Combination Sum III

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]]

这题和前面两题很像，但是限制了个数，而且不能重复，也就是不能出现[1,1,6]这种组合。因为candidates里没有重复数字，所以不用判断重复也不会出现重复的组合，如果candidates里有重复，要避免重复就使用注释的语句去重。

代码如下：

class Solution {
public:
    void dfs(vector<int> candidates, int start, int curtarget, int num, vector<int>& curcom, vector<vector<int>>& result) {
        if (curcom.size() == num && curtarget == 0) {
            result.push_back(curcom);
            return ;
        }
        if (curtarget < 0 || curcom.size() > num) return ;
        
        for (int i = start; i < candidates.size(); i++) {
            //if (i != candidates.size() && candidates[i] == candidates[i+1]) continue;
            curcom.push_back(candidates[i]);
            dfs(candidates, i + 1, curtarget - candidates[i], num, curcom, result);
            curcom.pop_back();
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<int> candidates = {1, 2, 3,4, 5, 6, 7, 8, 9};
        vector<vector<int>> result;
        vector<int> curcom;
        dfs(candidates, 0, n, k, curcom, result);
        return result;
    }
};