[LeetCode] 360. Sort Transformed Array

Given a sorted integer array nums and three integers a, b and c, apply a quadratic function of the form f(x) = ax2 + bx + c to each element nums[i] in the array, and return the array in a sorted order.

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]

Constraints:

• 1 <= nums.length <= 200
• -100 <= nums[i], a, b, c <= 100
• nums is sorted in ascending order.

Follow up: Could you solve it in O(n) time?

变换数组排序。

题意是给一个有序数组和几个数字a, b, c 请输出一个新的有序数组满足f(x) = ax^2 + bx + c。

这个题给的公式是初中学的一元二次方程。根据a的正负情况，函数的开口可上可下。这个题的思路依然是two pointer夹逼。因为函数的关系，会导致input数组中最小的和最大的数字在output里面的值也是最小的或者是最大的（取决于a的值）。所以当左右指针从两侧开始遍历input数组的时候，需要判断a的正负情况。如果a为正数，计算结果需要从res的尾部开始摆放；如果a为负数，计算结果需要从res的头部开始摆放。同时根据nums[start]和nums[end]的计算结果，决定到底是start++还是end--。

时间O(n)

空间O(n) - output

JavaScript实现

var sortTransformedArray = function (nums, a, b, c) {
    let res = new Array(nums.length);
    let start = 0;
    let end = nums.length - 1;
    let i = a >= 0 ? nums.length - 1 : 0;
    while (start <= end) {
        let startNum = getNum(nums[start], a, b, c);
        let endNum = getNum(nums[end], a, b, c);
        if (a >= 0) {
            if (startNum >= endNum) {
                res[i--] = startNum;
                start++;
            } else {
                res[i--] = endNum;
                end--;
            }
        } else {
            if (startNum <= endNum) {
                res[i++] = startNum;
                start++;
            } else {
                res[i++] = endNum;
                end--;
            }
        }
    }
    return res;
};

var getNum = function (x, a, b, c) {
    return a * x * x + b * x + c;
}

Java实现

class Solution {
    public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
        int[] res = new int[nums.length];
        int start = 0;
        int end = nums.length - 1;
        int i = a >= 0 ? nums.length - 1 : 0;
        while (start <= end) {
            int startNum = getNum(nums[start], a, b, c);
            int endNum = getNum(nums[end], a, b, c);
            if (a >= 0) {
                if (startNum >= endNum) {
                    res[i--] = startNum;
                    start++;
                } else {
                    res[i--] = endNum;
                    end--;
                }
            } else {
                if (startNum <= endNum) {
                    res[i++] = startNum;
                    start++;
                } else {
                    res[i++] = endNum;
                    end--;
                }
            }
        }
        return res;
    }

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

LeetCode 题目总结