LeetCode_Combination Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/leileicaocao/article/details/9206713

本博客介绍了一个算法，用于找出所有可能的候选数组合，这些组合的总和等于给定的目标值。允许重复选择候选数，并且组合必须按非降序排列，确保不包含重复解决方案。

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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, � , 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]

vector<vector<int> > combinationSum(vector<int> &arr, int target) 
{
    vector<vector<int> > ans;
	int size = arr.size();
	if (size == 0)
		return ans;

	sort(arr.begin(), arr.end());
	vector<pair<int, vector<int> > > vec;
	int *indexArr = new int[size];
	for (int i = 0; i < size; ++i)
	{
		indexArr[i] = i;
		vector<int> tmp;
		tmp.push_back(arr[i]);
		if (arr[i] == target)
		{
			ans.push_back(tmp);
		}
		else
		{
			vec.push_back(make_pair(arr[i], tmp));
		}
	}
	int end = vec.size();
	int index = 0;
	while (index < vec.size())
	{
		for (int i = 0; i < size; ++i)
		{
			int curNum = arr[i];
			int from = indexArr[i];
			indexArr[i] = vec.size();
			// 如果是最后一个，就到第0个
			int to = end;
			for (int j = from; j < to; ++j)
			{
				int sum = vec[j].first;
				++index;
				vector<int> tmp = vec[j].second;				
				if (sum + curNum <= target)
				{					
					tmp.push_back(curNum);
					if (sum + curNum == target)
					{
						sort(tmp.begin(), tmp.end());
						ans.push_back(tmp);
					}
					else
					{
						vec.push_back(make_pair(sum + curNum, tmp));
					}
				}
			}
		}
		index = end;
		end = vec.size();
	}
	return ans;
}