3Sum
<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>
Array; Two Pointers; Sorting
int cmp(const void* a, const void* b) {
    return *(int*) a - *(int*) b;
}

int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    int cap = 1000;
    int** ans = (int**) malloc(sizeof(int*) * cap);
    *returnColumnSizes = (int*) malloc(sizeof(int) * cap);

    qsort(nums, numsSize, sizeof(int), cmp);

    for (int i = 0; i < numsSize - 2 && nums[i] <= 0; ++i) {
        if (i > 0 && nums[i] == nums[i - 1]) continue;
        int j = i + 1, k = numsSize - 1;
        while (j < k) {
            int sum = nums[i] + nums[j] + nums[k];
            if (sum < 0) {
                ++j;
            } else if (sum > 0) {
                --k;
            } else {
                if (*returnSize >= cap) {
                    cap *= 2;
                    ans = (int**) realloc(ans, sizeof(int*) * cap);
                    *returnColumnSizes = (int*) realloc(*returnColumnSizes, sizeof(int) * cap);
                }
                ans[*returnSize] = (int*) malloc(sizeof(int) * 3);
                ans[*returnSize][0] = nums[i];
                ans[*returnSize][1] = nums[j];
                ans[*returnSize][2] = nums[k];
                (*returnColumnSizes)[*returnSize] = 3;
                (*returnSize)++;

                ++j;
                --k;
                while (j < k && nums[j] == nums[j - 1]) ++j;
                while (j < k && nums[k] == nums[k + 1]) --k;
            }
        }
    }
    return ans;
}
