leetcode-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)

思路：

     与3sum类似，先确定第一个，再确定第二个，另外两个用2sum方法确定.

代码：

vector<vector<int> > fourSum(vector<int> &num, int target) 
{
vector<vector<int>> res;
int len=num.size();
if(len<4)
{
return res;
}
sort(num.begin(),num.end());


for(int i=0;i<len-3;++i)
{
if(i!=0 && num[i]==num[i-1])
{
continue;
}
for(int j=i+1;j<len-2;++j)
{
if(j!=(i+1) && num[j]==num[j-1])
{
continue;
}
int p=j+1;
int q=len-1;
while(p<q)
{
if(num[i]+num[j]+num[p]+num[q] == target)
{
vector<int>triplet;
triplet.push_back(num[i]);
triplet.push_back(num[j]);
triplet.push_back(num[p]);
triplet.push_back(num[q]);
res.push_back(triplet);
while (++p < q  && num[p-1] == num[p])  
{  
}  
while (--q > p && num[q+1] == num[q])  
{  
}
} 
else if(num[i]+num[j]+num[p]+num[q] > target)
{
--q;
}
else
{
++p;
}
}
}
}
return res;
}