<p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>

<p>Notice that the solution set must not contain duplicate triplets.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
<strong>Explanation:</strong> 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,1]
<strong>Output:</strong> []
<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [[0,0,0]]
<strong>Explanation:</strong> The only possible triplet sums up to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
private:
    vector<int> nums;

    vector<vector<int>> twoSum(int begin, int end, int target, vector<int> nums) {
        int left = begin;
        int right = end;
        int base = nums[left] + nums[right];
        vector<vector<int>> sol_set;

        while (left < right) {
            if (base < target) {
                left++;
            } else if (base > target) {
                right--;
            } else {
                sol_set.push_back({nums[left], nums[right]});
                left++;
                right--;
                while (left < right && nums[left] == nums[left - 1]) left++;
            }
            base = nums[left] + nums[right];
        }

        return sol_set;
    }

public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;

        sort(nums.begin(), nums.end());
        int base = nums[0];
        for (int i = 0; i < nums.size() - 2; i++) {
            if (i > 0 && nums[i] == base) {
                continue;
            }
            vector<vector<int>> sub_res = twoSum(i + 1, nums.size() - 1, 0 - nums[i], nums);
            if (sub_res.empty()) {
                continue;
            }
            for (auto s : sub_res) {
                s.push_back(nums[i]);
                res.push_back(s);
            }
            base = nums[i];
        }

        return res;
    }
};
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