LeetCode 360. Sort Transformed Array

/*
  Given a sorted array of integers nums and integer values a, b and c.
  Apply a function of the form f(x) = ax^2 + bx + c to each element x in
  the array. The return array must be in sorted order.
  Expected time complexity: O(n)
  example:
  nums = [-4, -2, 2, 4], a = 1, b = 3, c = 5
  result = [3, 9, 15, 33]
*/
#include <vector>
#include <climits>
#include <algorithm>
#include <iostream>
using namespace std;

vector<int> sortTransformedArray(vector<int>& nums, int a, int b, int c) {
  vector<int> res;
  for(int i = 0; i < nums.size(); ++i) {
    int sum = a * nums[i] * nums[i] + b * nums[i] + c;
    res.push_back(sum);
  }
  sort(res.begin(), res.end());
  return res;
}
// a variation of merge sort.
vector<int> sortTransformedArrayII(vector<int>& nums, int a, int b, int c) {
  vector<int> res;
  int minValue = INT_MAX;
  int minIndex = 0;
  for(int i = 0; i < nums.size(); ++i) {
    int tmp = a * nums[i] * nums[i] + b * nums[i] + c;
    if(tmp < minValue) {
      minValue = tmp;
      minIndex = i;
    }
  }
  int front = minIndex;
  int end = minIndex + 1;
  while(front >= 0 && end < nums.size()) {
    int front_value = a * nums[front] * nums[front] + b * nums[front] + c;
    int end_value = a * nums[end] * nums[end] + b * nums[end] + c;
    if(front_value < end_value) {
      res.push_back(front_value);
      front--;
    } else {
      res.push_back(end_value);
      end++;
    }
  }
  while(front >= 0) {
    int front_value = a * nums[front] * nums[front] + b * nums[front] + c;
    res.push_back(front_value);
    front--;
  }
  while(end < nums.size()) {
    int end_value = a * nums[end] * nums[end] + b * nums[end] + c;
    res.push_back(end_value);
    end++;
  }
  return res;
}

int main(void) {
  vector<int> nums {-4, -2, 2, 4};
  vector<int> res = sortTransformedArrayII(nums, -1, 3, 5);
  for(int i = 0; i < res.size(); ++i) {
    cout << res[i] << endl;
  }
}