[Leetcode] Sort Transformed Array 拋物線排序

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]

對撞指針

複雜度

O(N) 時間 O(N) 空間

思路

拋物線中點: -b/2a
分這麼幾種狀況：
a > 0，
a < 0,
a == 0 && b >= 0,
a == 0 && b < 0
給的x數組是排序的，搞兩個指針從兩邊比較

注意

維護一個nextIndex能夠免去用linkedlist從而節省時間

代碼

public class Solution {
    public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
        int[] result = new int[nums.length];
        int start = 0, end = nums.length - 1;
        int nextIndex = 0;
        if (a > 0 || (a == 0 && b >= 0)) 
            nextIndex = nums.length - 1;
        if (a < 0 || (a == 0 && b < 0))
            nextIndex = 0;
        double mid = -1 * ((b * 1.0)  / (2.0 * a));
        while (start <= end) {
            if (a > 0 || (a == 0 && b >= 0)) {
                if (Math.abs(mid - nums[start]) > Math.abs(nums[end] - mid)) {
                    int x = nums[start++];
                    result[nextIndex--] = a * x * x + b * x + c;
                }
                else {
                    int x = nums[end--];
                    result[nextIndex--] = a * x * x + b * x + c;
                }
            }
            else if (a < 0 || (a == 0 && b < 0)){
                if (Math.abs(mid - nums[start]) > Math.abs(nums[end] - mid)) {
                    int x = nums[start++];
                    result[nextIndex++] = a * x * x + b * x + c;
                }
                else {
                    int x = nums[end--];
                    result[nextIndex++] = a * x * x + b * x + c;
                }
            }
        }
        return result;
    }
}