<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>> threeSum(vector<int>& nums) {
        int i = 0;
        int j = nums.size() - 1;
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        while(i<j-1){
            int tmp = 0 - nums[i];
            int k = i + 1;
            while(k<j){
            if(nums[k]+nums[j]==tmp){
                vector<int> tmp1 = {nums[i],nums[k],nums[j]};
                bool valid = true;
                for(int x = 0; x < res.size(); x++){
                    if(res[x][0]==tmp1[0]&&res[x][1]==tmp1[1]&&res[x][2]==tmp1[2]){
                        valid = false;
                        break;
                    }
                }
                if(valid){
                    res.push_back(tmp1);
                }
                k++;
                j--;
            }
            else if(nums[k]+nums[j]>tmp){
                j--;
            }
            else{
                k++;
            }
            }
            i++;
            j = nums.size() - 1;
        }
        return res;
    }
};
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