Combination Sum II

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本博客介绍了一个算法，用于从给定的候选数集合中找出所有可能的组合，使得这些组合的元素之和等于指定的目标值。每个候选数只能在组合中使用一次，且组合中的数必须按非降序排列，确保不包含重复组合。

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

#include<iostream>
#include<vector>
#include<algorithm>
#include<numeric>
#include<set>
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)
	{
		if (candidates[i] == candidates[i - 1] && i>num)//保证不前一个重复数字必须被使用
			continue;
		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 + 1);
		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;
}