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1685. 有序数组中差绝对值之和

English Version

题目描述

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 <= 105
• 1 <= nums[i] <= nums[i + 1] <= 104

解法

前缀和实现。

Python3

class Solution:
    def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:
        n = len(nums)
        presum = [0] * (n + 1)
        for i in range(n):
            presum[i + 1] = presum[i] + nums[i]
        res = []
        for i, num in enumerate(nums):
            t = num * i - presum[i] + presum[n] - presum[i + 1] - num * (n - i - 1)
            res.append(t)
        return res

Java

class Solution {
    public int[] getSumAbsoluteDifferences(int[] nums) {
        int n = nums.length;
        int[] presum = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            presum[i + 1] = presum[i] + nums[i];
        }
        int[] res = new int[n];
        for (int i = 0; i < n; ++i) {
            res[i] = nums[i] * i - presum[i] + presum[n] - presum[i + 1] - nums[i] * (n - i - 1);
        }
        return res;
    }
}

C++

class Solution {
public:
    vector<int> getSumAbsoluteDifferences(vector<int>& nums) {
        int n = nums.size();
        vector<int> presum(n + 1);
        for (int i = 0; i < n; ++i) {
            presum[i + 1] = presum[i] + nums[i];
        }
        vector<int> res;
        for (int i = 0; i < n; ++i) {
            int t = nums[i] * i - presum[i] + presum[n] - presum[i + 1] - nums[i] * (n - i - 1);
            res.push_back(t);
        }
        return res;
    }
};

Go

func getSumAbsoluteDifferences(nums []int) []int {
	n := len(nums)
	presum := make([]int, n+1)
	for i := 0; i < n; i++ {
		presum[i+1] = presum[i] + nums[i]
	}
	var res []int
	for i := 0; i < n; i++ {
		t := nums[i]*i - presum[i] + presum[n] - presum[i+1] - nums[i]*(n-i-1)
		res = append(res, t)
	}
	return res
}

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