<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) {
        unordered_map<int,vector<int>> targets;
        for (int i = 0; i < nums.size(); i++) {
            targets[nums[i]].push_back(i);
        }

        vector<vector<int>> result;
        int last_ni = INT32_MIN;
        int last_nj = INT32_MIN;
        for (int i = 0; i < nums.size(); i++) {
            for (int j = i + 1; j < nums.size(); j++) {
                if ((last_ni == nums[i] && last_nj == nums[j]) || (last_ni == nums[j] && last_nj == nums[i])) {
                    continue;
                }
                last_ni = nums[i];
                last_nj = nums[j];
                int t = -(nums[i] + nums[j]);
                for (int k : targets[t]) {
                    if (i != k && j != k) {
                        vector<int> triplet = {nums[i], nums[j], nums[k]};
                        sort(triplet.begin(), triplet.end());
                        result.emplace_back(triplet);
                        break;
                    }
                }
            }
        }

        if (result.size()) {
            sort(result.begin(), result.end());
            int w = 1;
            for (int r = 1; r < result.size(); r++) {
                auto& a = result[w - 1];
                auto& b = result[r];
                if (a[0] != b[0] || a[1] != b[1] || a[2] != b[2]) {
                    if (w != r) {
                        result[w] = move(result[r]);
                    }
                    w++;
                }
            }
            result.resize(w);
        }
        return result;
    }
};
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