3Sum Closest
<p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>

<p>Return <em>the sum of the three integers</em>.</p>

<p>You may assume that each input would have exactly one solution.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,2,1,-4], target = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0], target = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> The sum that is closest to the target is 0. (0 + 0 + 0 = 0).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 500</code></li>
	<li><code>-1000 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>-10<sup>4</sup> &lt;= target &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Two Pointers; Sorting
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int ans = 1 << 30;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int j = i + 1, k = n - 1;
            while (j < k) {
                int t = nums[i] + nums[j] + nums[k];
                if (t == target) return t;
                if (abs(t - target) < abs(ans - target)) ans = t;
                if (t > target)
                    --k;
                else
                    ++j;
            }
        }
        return ans;
    }
};