leetcode 18. 4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

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>> fourSum(vector<int>& nums, int target)
	{
		if (nums.empty())
			return re;
		map<int, int>count;
		for (int i = 0; i < nums.size(); i++)
			count[nums[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 neednum = 4;
		//答案里只有一种数字  
		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  
			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

其实哥的这个代码实用性很强，神马4sum，5sum。。。照单全收，neednum一改，一撸到底，卧槽，有没有。

然后往下一看，额，来了个3sum，好吧，copy、paste、accepted，尼玛，这才是复用啊，根本不用改的。

（请忽略以上chui扯niu淡bi）