# 360. Sort Transformed Array

Given a **sorted** array of integers nums and integer values a, b and c. Apply a quadratic 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 1:**

```
Input: nums = [-4,-2,2,4], a = 1, b = 3, c = 5
Output: [3,9,15,33]
```

**Example 2:**

```
Input: nums = [-4,-2,2,4], a = -1, b = 3, c = 5
Output: [-23,-5,1,7]
```

## Thoughts

排好序的数组问对每个数做ax^2 + bx + c后结果的排序。二次方的抛物线分为开头向上和向下。 用两个指针分别指向首尾，当开头向上意味着两端要比中心大，从后往前插入结果，插谁取决于两个指针代表的数谁大；开头向下类似，方向倒过来。

## Code

```python
class Solution:
    def sortTransformedArray(self, nums: List[int], a: int, b: int, c: int) -> List[int]:
        def cal(x):
            return a * x ** 2 + b * x + c
        i, j, res = 0, len(nums) - 1, []
        while i <= j:
            p, q = cal(nums[i]), cal(nums[j])
            if p > q and a > 0 or p < q and a <= 0:
                i += 1
                res.append(p)
            else: 
                j -= 1
                res.append(q)
        if a > 0: res.reverse()
        return res
        
```

```java
class Solution {
    public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
        int[] res = new int[nums.length];
        int l = 0, r = nums.length - 1;
        int index = a > 0 ? nums.length - 1 : 0;
        while (l <= r) {
            int left = quad(nums[l], a, b, c), right = quad(nums[r], a, b, c);
            if (a > 0) {
                if (left > right) {
                    res[index--] = left;
                    l++;
                } else {
                    res[index--] = right;
                    r--;
                }
            } else {
                if (left < right) {
                    res[index++] = left;
                    l++;
                } else {
                    res[index++] = right;
                    r--;
                }
            }
        }
        return res;
    }

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

## Analysis

时间复杂度O(N).


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://hao-fu-1.gitbook.io/oj/multiple-pointers/two_sum/sort-transformed-array.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.