Initially, you have a bank account balance of 100 dollars.
You are given an integer purchaseAmount representing the amount you will spend on a purchase in dollars.
At the store where you will make the purchase, the purchase amount is rounded to the nearest multiple of 10. In other words, you pay a non-negative amount, roundedAmount, such that roundedAmount is a multiple of 10 and abs(roundedAmount - purchaseAmount) is minimized.
If there is more than one nearest multiple of 10, the largest multiple is chosen.
Return an integer denoting your account balance after making a purchase worth purchaseAmount dollars from the store.
Note: 0 is considered to be a multiple of 10 in this problem.
\n
Example 1:
\n\n\nInput: purchaseAmount = 9\nOutput: 90\nExplanation: In this example, the nearest multiple of 10 to 9 is 10. Hence, your account balance becomes 100 - 10 = 90.\n\n\n
Example 2:
\n\n\nInput: purchaseAmount = 15\nOutput: 80\nExplanation: In this example, there are two nearest multiples of 10 to 15: 10 and 20. So, the larger multiple, 20, is chosen.\nHence, your account balance becomes 100 - 20 = 80.\n\n\n
\n
Constraints:
\n\n0 <= purchaseAmount <= 100一开始，你的银行账户里有 100 块钱。
给你一个整数purchaseAmount ，它表示你在一次购买中愿意支出的金额。
在一个商店里，你进行一次购买，实际支出的金额会向 最近 的 10 的 倍数 取整。换句话说，你实际会支付一个 非负 金额 roundedAmount ，满足 roundedAmount 是 10 的倍数且 abs(roundedAmount - purchaseAmount) 的值 最小 。
如果存在多于一个最接近的 10 的倍数，较大的倍数 是你的实际支出金额。
请你返回一个整数，表示你在愿意支出金额为 purchaseAmount 块钱的前提下，购买之后剩下的余额。
注意： 0 也是 10 的倍数。
\n\n
示例 1：
\n\n输入：purchaseAmount = 9\n输出：90\n解释：这个例子中，最接近 9 的 10 的倍数是 10 。所以你的账户余额为 100 - 10 = 90 。\n\n\n
示例 2：
\n\n输入：purchaseAmount = 15\n输出：80\n解释：这个例子中，有 2 个最接近 15 的 10 的倍数：10 和 20，较大的数 20 是你的实际开销。\n所以你的账户余额为 100 - 20 = 80 。\n\n\n
\n\n
提示：
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