Leetcode: 3sum closet

Question: For an array and a given target value, it is required to find three elements in the array. The sum of the three elements is the closest to the target value. Of course, it is the best.

Use the 3sum method to modify the condition.

int twoSum(vector
 
   &num, int start, int target){if(num.size() < 3 || start >= num.size())return -target;int head = start;int tail = num.size() - 1;int result = -target;int erro = 0x7fffffff;while (head < tail){int sum = num[head] + num[tail];if(sum == target){result = target;return target;    }else if(sum < target){do {++head;} while (head < tail && num[head] == num[head - 1]);}else{do {--tail;} while (tail > head && num[tail] == num[tail + 1]);}int curerro = abs(sum - target);if(curerro < erro){result = sum;erro = curerro;}}return result;}int threeSumClosest(vector
  
    &num, int target) {const int len = num.size();if(len < 3)return 0;sort(num.begin(), num.end());int closet = -target;int erro = 0x7fffffff;for (int i = 0; i < len - 2; ++i){int thisloop = twoSum(num, i + 1, target - num[i]);if(thisloop == target - num[i])return target;int cur_erro = abs(thisloop + num[i] - target);    if(cur_erro < erro)  {  closet = thisloop + num[i];  erro = cur_erro;}while(i + 1 < len && num[i + 1] == num[i]) ++i;}return closet;}