leetcode 40. Combination Sum II

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

class Solution {
	struct Info
	{
		vector<pair<int, int>>pre;//pair<int,int>里的first代表sinnums里的index,second代表个数    
		int remainsize;//剩余的数字个数    
		int remainvalue;//剩余的值    
	};
	vector<vector<int>>re;
	void choose_one(int&k, vector<Info>&candi, map<int, int>&count, vector<int>sinnums)
	{
		vector<Info>newcandi;
		if (k == 1)
		{
			for (int i = 0; i < candi.size(); i++)//考虑0    
				if (candi[i].remainvalue%candi[i].remainsize == 0)
				{
					int nxtnum = candi[i].remainvalue / candi[i].remainsize;
					if (nxtnum>sinnums[candi[i].pre.back().first] && count.find(nxtnum) != count.end()
						&& count[nxtnum] >= candi[i].remainsize)
					{
						vector<int>aa;
						for (int j = 0; j < candi[i].pre.size(); j++)
							for (int h = 0; h < candi[i].pre[j].second; h++)
								aa.push_back(sinnums[candi[i].pre[j].first]);
						for (int j = 0; j < candi[i].remainsize; j++)
							aa.push_back(nxtnum);
						re.push_back(aa);
					}
				}
			k--;
			return;
		}
		else
		{
			for (int i = 0; i < candi.size(); i++)
			{
				for (int j = 1; j <= (candi[i].remainsize - k + 1); j++)
				{
					for (int h = candi[i].pre.back().first + 1; (h < sinnums.size() - k + 1); h++)
					{
						//粗略做判断，减小候选集大小    
						if (candi[i].remainvalue <= sinnums[h] * j + sinnums.back()* (candi[i].remainsize - j)
							&& candi[i].remainvalue >= sinnums[h] * j + sinnums[h + 1] * (candi[i].remainsize - j)
							&& j <= count[sinnums[h]])
						{
							Info info;
							info.pre = candi[i].pre;
							info.pre.push_back(pair<int, int>(h, j));
							info.remainsize = candi[i].remainsize - j;
							info.remainvalue = candi[i].remainvalue - j*sinnums[h];
							newcandi.push_back(info);
						}
					}
				}
			}
			k--;
			candi = newcandi;
			return;
		}
	}

public:
	vector<vector<int>> combinationSum2(vector<int>& candidates, int target)
	{
		if (candidates.empty())
			return re;
		map<int, int>count;
		for (int i = 0; i < candidates.size(); i++)
			count[candidates[i]]++;
		map<int, int>::iterator it = count.end(); it--;
		if (target >= 0 && count.begin()->first> target
			|| target < 0 && it->first < target)
			return re;
		vector<int>sinnums;
		for (it = count.begin(); it != count.end(); it++)
			sinnums.push_back(it->first);
		int maxneednum = target/sinnums.front();
		int neednum;
		//答案里只有一种数字 
		for (neednum = 1; neednum <= maxneednum;neednum++)
		if (target%neednum == 0 && count.find(target / neednum) != count.end() && count[target / neednum] >= neednum)
		{
			vector<int>aa(neednum, target / neednum);
			re.push_back(aa);
		}
		for (int i = 0; i < sinnums.size(); i++)
		{
			//答案里有k种数字,2<=k<=neednum 
			int need = target / sinnums[i];
			if (need >= 2)
			{
				for (neednum = 2; neednum <= need; neednum++)
					for (int k = 2; k <= neednum; k++)
					{
						if (sinnums.size() - i >= k)
						{
							int jjmax = neednum - k + 1 < count[sinnums[i]] ? neednum - k + 1 : count[sinnums[i]];
							for (int j = 1; j <= jjmax; j++)
							{
								Info inf;
								inf.pre.push_back(pair<int, int>(i, j));
								inf.remainsize = neednum - j;
								inf.remainvalue = target - j*sinnums[i];
								vector<Info>candi;
								candi.push_back(inf);
								int kk = k - 1;
								while (kk > 0)
									choose_one(kk, candi, count, sinnums);
							}
						}
					}
			}
		}
		return re;
	}
};

accepted