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

**Input:** nums = [0,0,0], target = 1
**Output:** 0
**Explanation:** The sum that is closest to the target is 0. (0 + 0 + 0 = 0). 

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int closest = nums[0] + nums[1] + nums[2], n = nums.size();
        int diff = abs(closest - target);
        sort(nums.begin(), nums.end());
        for (int i = 0; i < n - 2; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int left = i + 1, right = n - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                int newDiff = abs(sum - target);
                if (diff > newDiff) {
                    diff = newDiff;
                    closest = sum;
                }
                if (sum < target) ++left;
                else --right;
            }
        }
        return closest;
    }
};