LeetCode 18: 4Sum

Difficulty: 3

Frequency: 2

Problem:

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)

Solution:

class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int> > answer;
if (num.size()<4)
return answer;

sort(num.begin(), num.end());

for (int i = num.size() - 1; i>2; --i)
{
int i_size = answer.size();
threeSum(num, i, answer, target - num[i]);
for (; i_size<answer.size(); i_size++)
{
answer[i_size].push_back(num[i]);
}
while(i>2&&num[i]==num[i-1])
--i;
}
return answer;
}
void threeSum(vector<int> &num, int size, vector<vector<int> > & answer, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (num.size()<3)
return;

vector<int> triplet;
int i_start, i_end;
int sum;
for (int i = 0; i < size -2; ++i)
{
if (i>0&&num[i]==num[i-1])
continue;

i_start = i+1;
i_end = size -1;

while (i_start<i_end)
{
if (num[i_start]==num[i_start+1]&&num[i] + num[i_start] + num[i_start+1]==target)
{
triplet.clear();
triplet.push_back(num[i]);
triplet.push_back(num[i_start]);
triplet.push_back(num[i_start]);
answer.push_back(triplet);
}
while (i_start<i_end&&num[i_start]==num[i_start+1])
++i_start;

if (i_start==i_end)
break;

if (num[i_end]==num[i_end-1]&&num[i] + num[i_end-1] + num[i_end]==target)
{
triplet.clear();
triplet.push_back(num[i]);
triplet.push_back(num[i_end]);
triplet.push_back(num[i_end]);
answer.push_back(triplet);
}
while (i_end>i_start&&num[i_end]==num[i_end-1])
--i_end;

sum = num[i] + num[i_start] + num[i_end];
if (sum==target)
{
triplet.clear();
triplet.push_back(num[i]);
triplet.push_back(num[i_start]);
triplet.push_back(num[i_end]);
answer.push_back(triplet);
++i_start;
}
else if (sum>target)
--i_end;
else
++i_start;
}
}
}
};

Notes:

Solution is based on 3Sum.