4 Sum
Given an array nums
of n
integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]]
such that:
0 <= a, b, c, d < n
a
,b
,c
, andd
are distinct.nums[a] + nums[b] + nums[c] + nums[d] == target
You may return the answer in any order.
Example 1:
Input: nums = [1,0,-1,0,-2,2], target = 0 Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Example 2:
Input: nums = [2,2,2,2,2], target = 8 Output: [[2,2,2,2]]
Constraints:
1 <= nums.length <= 200
-109 <= nums[i] <= 109
-109 <= target <= 109
Solution To the Above Question
class Solution
{
public:
vector<vector<int>> fourSum(vector<int> &nums, int target)
{
sort(nums.begin(), nums.end());
int n = nums.size();
// for(int i=0; i<n; i++){
// cout<<nums[i]<<" ";
// }
cout << endl;
vector<vector<int>> ans;
vector<int> v;
if (nums.size() < 4)
{
return ans;
}
for (int i = 0; i < n - 3; i++)
{
for (int x = i + 1; x < n - 2; x++)
{
int j = x + 1;
int k = n - 1;
while (j < k)
{
long long int current = nums[i];
current += nums[x];
current += nums[j];
current += nums[k];
// cout<<current<<" ";
if (current == target)
{
v.push_back(nums[i]);
v.push_back(nums[x]);
v.push_back(nums[j]);
v.push_back(nums[k]);
bool flag = false;
for (int m = 0; m < ans.size(); m++)
{
if (ans[m] == v)
{
flag = true;
}
}
if (!flag)
{
ans.push_back(v);
}
v.clear();
j++;
}
else if (current > target)
{
k--;
}
else if (current < target)
{
j++;
}
}
}
}
return ans;
}
};
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