Combination Sum

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本文介绍了一种基于递归的组合求和算法，用于找出所有可能的候选数组合，使得这些候选数的总和等于目标值。算法确保了组合中的元素非递减排序且不包含重复组合。

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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
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]

解题思路:用递归实现,题目要求：(a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).首先将列表排序.递归时下一个深度级的遍历开始位置由上一个深度级决定。用这样的方式保证要求.

#include<iostream>
#include<vector>
#include<algorithm>
#include<numeric>
using namespace std;

void FindcombinationSum(vector<vector<int> >&ResultVector, vector<int> &candidates, vector<int> &Oneresult, int target, int num = 0)
{
	if (accumulate(Oneresult.begin(), Oneresult.end(), 0) >= target)
		return;
	for (int i = num; i != candidates.size();++i)
	{
		Oneresult.push_back(candidates[i]);
		int sum = accumulate(Oneresult.begin(),Oneresult.end(),0);
		if (sum == target)
			ResultVector.push_back(Oneresult);
		FindcombinationSum(ResultVector, candidates, Oneresult, target, i);
		Oneresult.pop_back();
	}
}
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
	vector<vector<int> > ResultVector;
	vector<int>          Oneresult;
	sort(candidates.begin(), candidates.end());
	FindcombinationSum(ResultVector, candidates, Oneresult, target);
	return ResultVector;
}