【leetcode】1685. Sum of Absolute Differences in a Sorted Array

题目如下：

You are given an integer array nums sorted in non-decreasing order.

Build and return an integer array result with the same length as nums such that result[i] is equal to the summation of absolute differences between nums[i] and all the other elements in the array.

In other words, result[i] is equal to sum(|nums[i]-nums[j]|) where 0 <= j < nums.length and j != i (0-indexed).

Example 1:

Input: nums = [2,3,5]
Output: [4,3,5]
Explanation: Assuming the arrays are 0-indexed, then
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.

Example 2:

Input: nums = [1,4,6,8,10]
Output: [24,15,13,15,21]

Constraints:

• 2 <= nums.length <= 105
• 1 <= nums[i] <= nums[i + 1] <= 104

解题思路：对于nums[i]，我们很容易可以求出0~i-1区间和total_left以及i+1~nums.length-1的区间和total_right，那么对于左半部分的sum有：i*nums[i] - total_left，右半部分有：total_right - (len(nums) - i - 1)*nums[i]。

代码如下：

class Solution(object):
    def getSumAbsoluteDifferences(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        total = []
        count = 0
        for i in nums:
            count += i
            total.append(count)

        res = []
        for i in range(0,len(nums)):
            left = right = 0
            if i > 0:
                left = i*nums[i] - total[i-1]
            if i < len(nums) - 1:
                right = (total[-1] - total[i]) - (len(nums) - i - 1)*nums[i]
            res.append(left + right)
        return res