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]

Solution：Two pointers

思路： two ends 向内依次最大(a>=0) 或最小(a<0)

屏幕快照 2017-09-10 下午11.59.24.png

Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

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;
    }
}