地址：https://leetcode.com/problems/combination-sum/

題目：

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Example 2:

理解：

本題需要尋找給定集合中哪些子集的和等於target。雖然原來的集合沒有重複元素，但是元素是可以重複利用的。
基本思想，取集合中的一個元素candidates[i]，然後判斷剩下的集合元素能否構成target-candidates[i]。注意剩下的集合是可以包含candidates[i]

實現：

Accepted 16ms c++ solution use backtracking, easy understand.
這種實現使用排序避開了需要遍歷完整個集合。

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