左右元素和的差值

Given a 0-indexed integer array nums, find a 0-indexed integer array answer where:

• answer.length == nums.length.
• answer[i] = |leftSum[i] - rightSum[i]|.

Where:

• leftSum[i] is the sum of elements to the left of the index i in the array nums. If there is no such element, leftSum[i] = 0.
• rightSum[i] is the sum of elements to the right of the index i in the array nums. If there is no such element, rightSum[i] = 0.

Return the array answer.

 

Example 1:

Input: nums = [10,4,8,3]
Output: [15,1,11,22]
Explanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].
The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].

Example 2:

Input: nums = [1]
Output: [0]
Explanation: The array leftSum is [0] and the array rightSum is [0].
The array answer is [|0 - 0|] = [0].

 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 105

/*
 * 题目思路：
 * 通过使用 Java Stream 来计算 leftSum 和 rightSum。
 * 我们首先使用 Stream 来生成 leftSum 和 rightSum，
 * 然后用 mapToInt 方法计算绝对值差。
 */
import java.util.stream.*;

public class Solution {
    public int[] leftRightDifference(int[] nums) {
        int n = nums.length;
        int[] leftSum = new int[n];
        int[] rightSum = new int[n];

        // 计算 leftSum 数组
        IntStream.range(1, n).forEach(i -> leftSum[i] = leftSum[i - 1] + nums[i - 1]);

        // 计算 rightSum 数组
        IntStream.iterate(n - 2, i -> i >= 0, i -> i - 1).forEach(i -> rightSum[i] = rightSum[i + 1] + nums[i + 1]);

        // 计算 answer 数组
        return IntStream.range(0, n)
                .map(i -> Math.abs(leftSum[i] - rightSum[i]))
                .toArray();
    }
}

题解采用一次遍历的方法计算左右和的差的绝对值。首先，初始化leftsum为0（因为第一个元素左侧无元素），rightsum为除了第一个元素外的所有元素之和。然后遍历数组，每到一个新的元素i，计算当前的leftsum和rightsum的差的绝对值，并更新leftsum和rightsum，以便在下一次循环时使用。leftsum每次增加当前元素nums[i]，而rightsum减去下一个元素nums[i+1]（这是因为该元素将从右侧和中移除）。最后，将最后一个元素的差的绝对值也添加到结果中。

O(n)

O(n)