Leetcode39. Combination Sum

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本文介绍了解决LeetCode第39题“组合总和”的方法，使用了回溯算法来找出所有可能的数列组合，这些组合中的元素之和等于目标值。代码实现中对输入进行了排序以提高效率，并详细展示了如何递归地添加和移除元素。

Leetcode39. Combination Sum

题目：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]
]

题意分析：这道题，选择用回溯的方法去做。采用的方法也是类似背包问题，先假设放进去，不行再取出来。

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();
	}
}

代码：

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();
		}
	}
};