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

Hide Hint 1

Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted?

Hide Hint 2

For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]).

Hide Hint 3

It can be simplified to (nums[i] * i - (nums[0] + nums[1] + ... + nums[i-1])) + ((nums[i+1] + nums[i+2] + ... + nums[n-1]) - nums[i] * (n-i-1)). One can build prefix and suffix sums to compute this quickly.

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直接按照提示一步一步算就可以了。下面是代碼：

Runtime:?4 ms, faster than?62.05%?of?Java?online submissions for?Sum of Absolute Differences in a Sorted Array.

Memory Usage:?57.5 MB, less than?77.44%?of?Java?online submissions for?Sum of Absolute Differences in a Sorted Array.