Combination Sum II

最新推荐文章于 2021-02-23 05:50:14 发布

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最新推荐文章于 2021-02-23 05:50:14 发布

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分类专栏： c++/c

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本文讨论了如何在给定候选数集合和目标数的情况下，找出所有满足条件的唯一组合，确保每个数只使用一次且组合有序，避免重复。通过引入01背包问题的概念并实施有效的剪枝策略来提高算法效率，最终实现目标数的精准匹配。包括对排序、剪枝逻辑的优化及测试用例分析，以验证算法的有效性和正确性。

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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]

利用01背包，另外减枝。注意减枝尽量多。

开始由于未加 if(target < 0 ) return; 结果导致严重超时，即使代码是对的。但是由于他任然在遍历所有可能，但是不退出。最小面给出测试用例，可以发现，超时即为严重！

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

void combine_sum(vector<int> &candidate,int target,int level ,vector<bool> &used,vector<int> res_one,vector<vector<int>> &res);
vector<vector<int>>  combine_sum_final(vector<int> &candidate,int target);

int main()
{

	vector<int> candidate;
	int target;
	int temp;
	cin >> temp;
	while (temp >= 0)
	{
		candidate.push_back(temp);
		cin >> temp;
	}
	cout<<endl;
	cin>>target;


	vector<vector<int>> res;
	res = combine_sum_final(candidate,target);
	cout<<"res:" <<res.size()<<endl;
	for (int i = 0 ;i < res.size();i++)
	{
		for (int j = 0;j <res[i].size();j++)
		{
			cout<< res[i][j]<<" ";
		}
		cout<<endl;
	}
}

vector<vector<int>>  combine_sum_final(vector<int> &candidate,int target)
{
	sort(candidate.begin(),candidate.end());//<span style="color:#cc0000;">先排序</span>
	if (target == 0 || candidate.size() == 0 || candidate[0] <= 0)
	{
		return vector<vector<int>>();
	}
	vector<int>  res_one;
	vector<vector<int>> res;
	vector<bool> used(candidate.size(),false);
	combine_sum(candidate,target,0,used,res_one,res);

	return res;
}

//需保证target 非0 ！！  01 背包加减枝
void combine_sum(vector<int> &candidate,int target,int level ,vector<bool> &used,vector<int> res_one,vector<vector<int>> &res)
{
	 if(target < 0 ) return;
	 if (level >= candidate.size())
	 {
		 if (target == 0)
		 {
			 res.push_back(res_one);
		 }
		 return ;
	 }

	 bool uesd_falg = true;
/*
	 for(int i = 0;i < level;i++)
	 {
		if (candidate[i] == candidate[level] && (!used[i]))
		{
			uesd_falg = false;
			break;
		}
	 }
*/  //对付未排序的
	for(int i = level-1;i >= 0 && candidate[i] == candidate[level];i--)//排序了的
	{
		if (!used[i])
		{
			uesd_falg = false;
		}
	}
	if(uesd_falg)//加入前判断
	{
		res_one.push_back(candidate[level]);
		used[level] =  true;
		combine_sum(candidate,target-candidate[level],level+1,used,res_one,res);
		res_one.pop_back();
		used[level] =  false;
	}

	
	combine_sum(candidate,target,level+1,used,res_one,res);//不加入分支
		

}

第二次：

 vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        
        return combine_sum_final(num,target);
    }
    vector<vector<int>>  combine_sum_final(vector<int> &candidate,int target)
    {
	sort(candidate.begin(),candidate.end());
	if (target == 0 || candidate.size() == 0 || candidate[0] <= 0)
	{
		return vector<vector<int>>();
	}
	vector<int>  res_one;
	vector<vector<int>> res;
	vector<bool> used(candidate.size(),false);
	combine_sum(candidate,target,0,used,res_one,res);

	return res;
    }

void combine_sum(vector<int> &candidate,int target,int level ,vector<bool> &used,vector<int> res_one,vector<vector<int>> &res)
{
	 if(target < 0 ) return;
	 if (level >= candidate.size())
	 {
		 if (target == 0)
		 {
			 res.push_back(res_one);
		 }
		 return ;
	 }

	 bool uesd_falg = true;

	//写法也是对的，后面想想发现没必要依次往前看，只需看前一个，应为如果连续几个111 重复，010，001这种情况在前面已减枝，
	//后面只需判断101此种
	if(level > 0 && candidate[level-1] == candidate[level] && !used[level-1]) uesd_falg = false;
	
	if(uesd_falg)//加入前判断
	{
		res_one.push_back(candidate[level]);
		used[level] =  true;
		combine_sum(candidate,target-candidate[level],level+1,used,res_one,res);
		res_one.pop_back();
		used[level] =  false;
	}

	
	combine_sum(candidate,target,level+1,used,res_one,res);
		

}

//10 1 2 7 6 1 5 -1
//12

//14 6 25 9 30 20 33 34 28 30 16 12 31 9 9 12 34 16 25 32 8 7 30 12 33 20 21 29 24 17 27 34 11 17 30 6 32 21 27 17 16 8 24 12 12 28 11 33 10 32 22 13 34 18 12 -1（用于结束输入）

// 结果：27