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）