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;
}
};