You are given a string expression representing a Lisp-like expression to return the integer value of.
\n\nThe syntax for these expressions is given as follows.
\n\n"(let v1 e1 v2 e2 ... vn en expr)", where let is always the string "let", then there are one or more pairs of alternating variables and expressions, meaning that the first variable v1 is assigned the value of the expression e1, the second variable v2 is assigned the value of the expression e2, and so on sequentially; and then the value of this let expression is the value of the expression expr."(add e1 e2)" where add is always the string "add", there are always two expressions e1, e2 and the result is the addition of the evaluation of e1 and the evaluation of e2."(mult e1 e2)" where mult is always the string "mult", there are always two expressions e1, e2 and the result is the multiplication of the evaluation of e1 and the evaluation of e2."add", "let", and "mult" are protected and will never be used as variable names.\n
Example 1:
\n\n\nInput: expression = "(let x 2 (mult x (let x 3 y 4 (add x y))))"\nOutput: 14\nExplanation: In the expression (add x y), when checking for the value of the variable x,\nwe check from the innermost scope to the outermost in the context of the variable we are trying to evaluate.\nSince x = 3 is found first, the value of x is 3.\n\n\n
Example 2:
\n\n\nInput: expression = "(let x 3 x 2 x)"\nOutput: 2\nExplanation: Assignment in let statements is processed sequentially.\n\n\n
Example 3:
\n\n\nInput: expression = "(let x 1 y 2 x (add x y) (add x y))"\nOutput: 5\nExplanation: The first (add x y) evaluates as 3, and is assigned to x.\nThe second (add x y) evaluates as 3+2 = 5.\n\n\n
\n
Constraints:
\n\n1 <= expression.length <= 2000expression.expression.给你一个类似 Lisp 语句的字符串表达式 expression，求出其计算结果。
表达式语法如下所示:
\n\n\"(let v1 e1 v2 e2 ... vn en expr)\" 的形式，其中 let 总是以字符串 \"let\"来表示，接下来会跟随一对或多对交替的变量和表达式，也就是说，第一个变量 v1被分配为表达式 e1 的值，第二个变量 v2 被分配为表达式 e2 的值，依次类推；最终 let 表达式的值为 expr表达式的值。\"(add e1 e2)\" ，其中 add 总是以字符串 \"add\" 来表示，该表达式总是包含两个表达式 e1、e2 ，最终结果是 e1 表达式的值与 e2 表达式的值之 和 。\"(mult e1 e2)\" ，其中 mult 总是以字符串 \"mult\" 表示，该表达式总是包含两个表达式 e1、e2，最终结果是 e1 表达式的值与 e2 表达式的值之 积 。\"add\" ，\"let\" ，\"mult\" 会被定义为 \"关键字\" ，不会用作变量名。示例 1：
\n\n\n输入：expression = \"(let x 2 (mult x (let x 3 y 4 (add x y))))\"\n输出：14\n解释：\n计算表达式 (add x y), 在检查变量 x 值时，\n在变量的上下文中由最内层作用域依次向外检查。\n首先找到 x = 3, 所以此处的 x 值是 3 。\n\n\n
示例 2：
\n\n\n输入：expression = \"(let x 3 x 2 x)\"\n输出：2\n解释：let 语句中的赋值运算按顺序处理即可。\n\n\n
示例 3：
\n\n\n输入：expression = \"(let x 1 y 2 x (add x y) (add x y))\"\n输出：5\n解释：\n第一个 (add x y) 计算结果是 3，并且将此值赋给了 x 。 \n第二个 (add x y) 计算结果是 3 + 2 = 5 。\n\n \n\n
提示：
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"style": "LEETCODE",
"exampleTestcases": "\"(let x 2 (mult x (let x 3 y 4 (add x y))))\"\n\"(let x 3 x 2 x)\"\n\"(let x 1 y 2 x (add x y) (add x y))\"",
"__typename": "QuestionNode"
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