Array (8): 3Sum Closest

describe:

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).

Analysis:
Sort first, then squeeze left and right, complexity O(n^2).

Code:

class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
         int result = 0;
         int min_gap = INT_MAX;
         sort(nums.begin(), nums.end());
         for (auto a = nums.begin(); a != prev(nums.end(), 2); ++a) {
              auto b = next(a);
              auto c = prev(nums.end());
              while (b < c) {
                           const int sum = *a + *b + *c;
                           const int gap = abs(sum - target);
                           if (gap < min_gap) {
                                   result = sum;
                                   min_gap = gap;
                            }
                           if (sum < target) ++b;
                           else --c;
                }
          }
        return result;
      }
};