LeetCode题目之8 Combination Sum II

最新推荐文章于 2020-02-12 03:09:08 发布

doublechen_it

最新推荐文章于 2020-02-12 03:09:08 发布

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分类专栏：数据结构与算法文章标签： numbers combinations ie c

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本文介绍了一个经典的算法问题：给定一组候选数和目标数，找出所有可能的组合，使得组合内的数相加等于目标数。文章详细解释了递归算法的具体实现，并提供了 C++ 代码示例。

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题目：

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]

解析：递归求法：

void CombinationSumII(vector<int> l, int start, vector<int> v, int sum, vector<vector<int> > & ret)
{
		
		if (sum==0){
			 ret.push_back(v);
			return;
		}
		if ((start==l.size())||sum<0)
		{
			return;
		}
		int prev = -1;
		for(int i=start; i< l.size();i++){
			if(l[i]!=prev)
			{
			v.push_back(l[i]);
			CombinationSumII(l,i+1,v,sum-l[i],ret);
			v.pop_back();
			prev=l[i];
			}
		}

}
vector<vector<int> > combinationSum2(vector<int> &num, int target){
	vector<vector<int> > ret ;
        vector<int> v ;
	sort(num.begin(),num.end());
        CombinationSumII(num, 0, v, target, ret);
	
        return ret;

			
}