The description of the problem is : https://leetcode.com/problems/3sum/#/description

**Generally**, The time complexity is O(n^2). Use for loop and squeeze rule to solve this problem.

**For boundary condition**: consider when the number in candidates is less than 3, then there is no solution.

The solution:

class Solution {

public:

vector<vector<int>> threeSum(vector<int>& nums) {

if (nums.size() < 3) return{};

vector<vector<int>> result;

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

for (int i = 0; i < nums.size() – 2; i++)

{

if (nums[i] > 0)

break;

if (i > 0 && nums[i] == nums[i – 1])

continue;

int left = i + 1;

int right = nums.size() – 1;

while (left < right)

{

if (nums[right] < 0)

break;

int tmp = nums[i] + nums[left] + nums[right];

if (tmp == 0)

{

result.push_back({ nums[i], nums[left], nums[right] });

while (left < right && nums[left] == nums[left + 1])

left++;

while (left < right && nums[right] == nums[right – 1])

right–;

++left;

–right;

}

else if (tmp < 0)

left++;

else

right–;

}

}

return result;

}

};

**Tips:**

How to remove the duplicate elements in vector?

result.erase(std::unique(result.begin(), result.end()), result.end());

Shit, I already do that all and I haven’t written any of the convos

down!