Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

Example 4:

Input: candidates = [1], target = 1
Output: [[1]]

Example 5:

Input: candidates = [1], target = 2
Output: [[1,1]]

Constraints:

1 <= candidates.length <= 30
1 <= candidates[i] <= 200
All elements of candidates are distinct.
1 <= target <= 500

Solution

Time complexity : O(nlog(n))
Space complexity : O(n)

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        helper(candidates, {}, 0, target);
        return ans;
    }
private:
    vector<vector<int>> ans;
    void helper(vector<int>& candidates, vector<int> cur, int start, int target) {
        if (0 == target) {
            ans.push_back(cur);
            return;
        }
        for (int i=start; i<candidates.size(); ++i) {
            int n = candidates[i];
            if (n > target) break;
            cur.push_back(n);
            helper(candidates, cur, i, target-n);
            cur.pop_back();
        }
    }
};

sort完之後確定為小→大。
用backtracking的方式，得到加總為target的組合。