Leetcode39. Combination Sum

Leetcode39. Combination Sum

題目：Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

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

For example, given candidate set [2, 3, 6, 7] and target 7, 
A solution set is: 

[
  [7],
  [2, 2, 3]
]

題意分析：這道題，選擇用回溯的方法去做。採用的方法也是類似揹包問題，先假設放進去，不行再取出來。

void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::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();
	}
}

代碼：

class Solution {
public:
	std::vector<std::vector<int> > combinationSum(std::vector<int> &candidates, int target) {
		std::sort(candidates.begin(), candidates.end());
		std::vector<std::vector<int> > res;
		std::vector<int> combination;
		combinationSum(candidates, target, res, combination, 0);
		return res;
	}
private:

	void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::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();
		}
	}
};