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:
\n\nleftSum[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.
\n
Example 1:
\n\n\nInput: nums = [10,4,8,3]\nOutput: [15,1,11,22]\nExplanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].\nThe array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].\n\n\n
Example 2:
\n\n\nInput: nums = [1]\nOutput: [0]\nExplanation: The array leftSum is [0] and the array rightSum is [0].\nThe array answer is [|0 - 0|] = [0].\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 10001 <= nums[i] <= 105给你一个下标从 0 开始的整数数组 nums ，请你找出一个下标从 0 开始的整数数组 answer ，其中：
answer.length == nums.lengthanswer[i] = |leftSum[i] - rightSum[i]|其中：
\n\nleftSum[i] 是数组 nums 中下标 i 左侧元素之和。如果不存在对应的元素，leftSum[i] = 0 。rightSum[i] 是数组 nums 中下标 i 右侧元素之和。如果不存在对应的元素，rightSum[i] = 0 。返回数组 answer 。
\n\n
示例 1：
\n\n输入：nums = [10,4,8,3]\n输出：[15,1,11,22]\n解释：数组 leftSum 为 [0,10,14,22] 且数组 rightSum 为 [15,11,3,0] 。\n数组 answer 为 [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22] 。\n\n\n
示例 2：
\n\n输入：nums = [1]\n输出：[0]\n解释：数组 leftSum 为 [0] 且数组 rightSum 为 [0] 。\n数组 answer 为 [|0 - 0|] = [0] 。\n\n\n
\n\n
提示：
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