<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 {
public:

    vector<vector<int>> twosum(vector<int>& nums, int start, int target){
        int tmp, l=start, r=nums.size()-1;
        vector<vector<int>> ans;
        while(l<r){
            tmp=nums[l]+nums[r];
            if(tmp>target)
                r--;
            else if(tmp==target){
                ans.push_back({nums[r], nums[l]});
                l++;
                r--;
            }
            else
                l++;
        }
        return ans;
    }

    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int target;
        set<vector<int>> all;
        for (int i=0; i<nums.size()-2; i++){
            if(i!=0 && nums[i]==nums[i-1])
                continue;
            target=-nums[i];
            vector<vector<int>> tmp=twosum(nums,i+1,target);
            for(int j=0; j<tmp.size(); j++){
                tmp[j].push_back(nums[i]);
                all.insert(tmp[j]);
            }
        }
        vector<vector<int>> ans(all.begin(),all.end());
        return ans;
    }
};
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