[leetcode 17] 4Sum

题目：

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：

1.与上上篇文章（3Sum）思路一样，无非是多加了一层循环；
2.算法简单，可直接见代码。

代码：

class Solution{
public:
vector<vector<int> > fourSum(vector<int> &num,int target){
vector<vector<int> > result;
if(num.size()==0)
return result;
sort(num.begin(),num.end());
for(int i=0;i!=num.size();++i)
{
if(i>0 && num[i]==num[i-1])
continue;
for(int j=i+1;j!=num.size();++j)
{
if(j>i+1 && num[j]==num[j-1])
continue;
int k=j+1,l=num.size()-1;
while(k<l)
{
if(k>j+1 && num[k]==num[k-1])
{
++k;
continue;
}
if(l<num.size()-1 && num[l]==num[l+1])
{
--l;
continue;
}
if(num[i]+num[j]+num[k]+num[l]-target<0)
++k;
else if(num[i]+num[j]+num[k]+num[l]-target>0)
--l;
else
{
vector<int> t;
t.push_back(num[i]);
t.push_back(num[j]);
t.push_back(num[k]);
t.push_back(num[l]);
result.push_back(t);
++k;--l;
}
}
}
}
return result;

}
};