3Sum Closest

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.

    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(num.begin(), num.end());
        int sum = num[0] + num[1] + num[2];
        int result = sum;
        int min_gap = abs(sum - target);
        for (int i = 0; i < num.size() - 2; ++i) {
            int low = i + 1;
            int high = num.size() - 1;
            while (low < high) {
                sum = num[i] + num[low] + num[high];
                int gap = sum - target;
                if (gap > 0) {
                    min_gap = min_gap < gap ? min_gap : gap;
                    if (min_gap == gap) {
                        result = sum;
                    }
                    --high;
                } else if (gap < 0) {
                    min_gap = min_gap < -gap ? min_gap : -gap;
                    if (min_gap == -gap) {
                        result = sum;
                    }
                    ++low;
                } else {
                    return target;
                }
            }
        }
        return result;
    }
};