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) = 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]

一刷
题解：
当a大于0时，(f(a)+f(b))/2 > f((a+b)/2)
当a小于0时，(f(a)+f(b))/2 < f((a+b)/2)

从数组首位两端开始寻找最值

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;
    }
}