Sort Transformed Array

Description

Given a sorted array of integers nums and integer values a, b and c. Apply a function of the form f(x) = ax2 + bx + c to each element x in the array.

The returned 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]

nums = [-4, -2, 2, 4], a = -1, b = 3, c = 5

Result: [-23, -5, 1, 7]

Hint

Train of Thought

一元二次方程, 一个抛物线, a>0,开口向上, a<0, 开口向下.
a>0, 对一串x, 一定可以找到最大值, 因为可以全在对称轴一侧或者在两侧, 
a<0, 一定可以找到最小值
用双指针, 如果a>0, 从最大值开始往下找, a<0, 从最小值往大找
a=0, 一定可以找到最大值, 因为是linear的

Code

public class Solution {
    public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
        int n = nums.length;
        int[] sorted = new int[n];
        int i = 0, j = n - 1;
        int index = a >= 0 ? n - 1 : 0;
        while (i <= j) {
            if (a >= 0) {
                sorted[index--] = quad(nums[i], a, b, c) >= quad(nums[j], a, b, c) ? quad(nums[i++], a, b, c) : quad(nums[j--], a, b, c);
            } else {
                sorted[index++] = quad(nums[i], a, b, c) >= quad(nums[j], a, b, c) ? quad(nums[j--], a, b, c) : quad(nums[i++], a, b, c);
            }
        }
        return sorted;
    }

    private int quad(int x, int a, int b, int c) {
        return a * x * x + b * x + c;
    }
}

Complexity