leetcode - Combination Sum

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Given a set of candidate numbers (C) 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.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• 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] 

class Solution 
{
public:
	std::vector<std::vector<int> > combinationSum(std::vector<int> &candidates, int target)
	{
		std::sort(candidates.begin(), candidates.end());
		dfs(candidates, 0,0, target);
		return res;
	}
private:
	std::vector<int> path;
	std::vector<std::vector<int>> res;
	// start 从candidates开始的位置, sum当前的和,target目标值,path存放满足条件的值,res 存放所有结果
	void dfs(std::vector<int> &candidates, int start, int sum, int target)
	{
		if(sum>target)//超出目标值，直接返回
			return ;
		if(sum == target)// 满足条件
		{
			res.push_back(path);
			return ;
		}
		for(int i=start; i< candidates.size(); i++)
		{
			path.push_back(candidates[i]);
			dfs(candidates, i, sum+candidates[i], target);
			path.pop_back();
		}
	}
};

leetcode - Combination Sum

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原文地址：http://blog.csdn.net/akibatakuya/article/details/40451099