有序数组中差绝对值之和

• https://leetcode.cn/problems/sum-of-absolute-differences-in-a-sorted-array/description/

描述

• 给你一个 非递减 有序整数数组 nums

• 请你建立并返回一个整数数组 result，它跟 nums 长度相同，且result[i] 等于 nums[i] 与数组中所有其他元素差的绝对值之和

• 换句话说， result[i] 等于 sum(|nums[i]-nums[j]|) ，其中 0 <= j < nums.length 且 j != i （下标从 0 开始）

示例 1

输入：nums = [2,3,5]
输出：[4,3,5]

解释：假设数组下标从 0 开始，那么

result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4，
result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3，
result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5

示例 2

输入：nums = [1,4,6,8,10]
输出：[24,15,13,15,21]

提示

• 2 <= nums.length <= $10^5$
• 1 <= nums[i] <= nums[i + 1] <= $10^4$

Typescript 版算法实现

1 ) 方案1：前缀和

function getSumAbsoluteDifferences(nums: number[]): number[] {const n = nums.length;let leftSum = 0;let rightSum = nums.reduce((acc, val) => acc + val, 0);const result: number[] = new Array(n);for (let i = 0; i < n; i++) {rightSum -= nums[i];result[i] = i * nums[i] - leftSum + rightSum - (n - i - 1) * nums[i];leftSum += nums[i];}return result;
}

2 ) 方案：另一种方案

function getSumAbsoluteDifferences(nums: number[]): number[] {// 计算前缀和数组const prefixSums: number[] = [0];for (let i = 0; i < nums.length; i++) {prefixSums.push(prefixSums[i] + nums[i]);}const ans: number[] = [];const n = nums.length;for (let i = 0; i < n; i++) {const x = nums[i];const left = i * x - prefixSums[i];const right = prefixSums[n] - prefixSums[i + 1] - x * (n - 1 - i);ans.push(left + right);}return ans;
}