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
use std::cmp::Ordering;

impl Solution {
    pub fn three_sum(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
        nums.sort();
        let n = nums.len();
        let mut res = vec![];
        let mut i = 0;
        while i < n - 2 && nums[i] <= 0 {
            let mut l = i + 1;
            let mut r = n - 1;
            while l < r {
                match (nums[i] + nums[l] + nums[r]).cmp(&0) {
                    Ordering::Less => {
                        l += 1;
                    }
                    Ordering::Greater => {
                        r -= 1;
                    }
                    Ordering::Equal => {
                        res.push(vec![nums[i], nums[l], nums[r]]);
                        l += 1;
                        r -= 1;
                        while l < n && nums[l] == nums[l - 1] {
                            l += 1;
                        }
                        while r > 0 && nums[r] == nums[r + 1] {
                            r -= 1;
                        }
                    }
                }
            }
            i += 1;
            while i < n - 2 && nums[i] == nums[i - 1] {
                i += 1;
            }
        }
        res
    }
}
