![\"\"](\"https://assets.leetcode.com/uploads/2022/02/22/ex1drawio.png\")
You are given a 0-indexed integer array nums, where nums[i] is a digit between 0 and 9 (inclusive).
The triangular sum of nums is the value of the only element present in nums after the following process terminates:
nums comprise of n elements. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n - 1.i, where 0 <= i < n - 1, assign the value of newNums[i] as (nums[i] + nums[i+1]) % 10, where % denotes modulo operator.nums with newNums.Return the triangular sum of nums.
\n
Example 1:
\n\n\nInput: nums = [1,2,3,4,5]\nOutput: 8\nExplanation:\nThe above diagram depicts the process from which we obtain the triangular sum of the array.\n\n
Example 2:
\n\n\nInput: nums = [5]\nOutput: 5\nExplanation:\nSince there is only one element in nums, the triangular sum is the value of that element itself.\n\n
\n
Constraints:
\n\n1 <= nums.length <= 10000 <= nums[i] <= 9给你一个下标从 0 开始的整数数组 nums ，其中 nums[i] 是 0 到 9 之间（两者都包含）的一个数字。
nums 的 三角和 是执行以下操作以后最后剩下元素的值：
nums 初始包含 n 个元素。如果 n == 1 ，终止 操作。否则，创建 一个新的下标从 0 开始的长度为 n - 1 的整数数组 newNums 。0 <= i < n - 1 的下标 i ，newNums[i] 赋值 为 (nums[i] + nums[i+1]) % 10 ，% 表示取余运算。newNums 替换 数组 nums 。请你返回 nums 的三角和。
\n\n
示例 1：
\n\n
\n输入：nums = [1,2,3,4,5]\n输出：8\n解释：\n上图展示了得到数组三角和的过程。\n\n
示例 2：
\n\n\n输入：nums = [5]\n输出：5\n解释：\n由于 nums 中只有一个元素，数组的三角和为这个元素自己。\n\n
\n\n
提示：
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