leetcode-problemset/leetcode-cn/originData/parse-lisp-expression.json

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            "title": "Parse Lisp Expression",
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            "content": "<p>You are given a string expression representing a Lisp-like expression to return the integer value of.</p>\n\n<p>The syntax for these expressions is given as follows.</p>\n\n<ul>\n\t<li>An expression is either an integer, let expression, add expression, mult expression, or an assigned variable. Expressions always evaluate to a single integer.</li>\n\t<li>(An integer could be positive or negative.)</li>\n\t<li>A let expression takes the form <code>&quot;(let v<sub>1</sub> e<sub>1</sub> v<sub>2</sub> e<sub>2</sub> ... v<sub>n</sub> e<sub>n</sub> expr)&quot;</code>, where let is always the string <code>&quot;let&quot;</code>, then there are one or more pairs of alternating variables and expressions, meaning that the first variable <code>v<sub>1</sub></code> is assigned the value of the expression <code>e<sub>1</sub></code>, the second variable <code>v<sub>2</sub></code> is assigned the value of the expression <code>e<sub>2</sub></code>, and so on sequentially; and then the value of this let expression is the value of the expression <code>expr</code>.</li>\n\t<li>An add expression takes the form <code>&quot;(add e<sub>1</sub> e<sub>2</sub>)&quot;</code> where add is always the string <code>&quot;add&quot;</code>, there are always two expressions <code>e<sub>1</sub></code>, <code>e<sub>2</sub></code> and the result is the addition of the evaluation of <code>e<sub>1</sub></code> and the evaluation of <code>e<sub>2</sub></code>.</li>\n\t<li>A mult expression takes the form <code>&quot;(mult e<sub>1</sub> e<sub>2</sub>)&quot;</code> where mult is always the string <code>&quot;mult&quot;</code>, there are always two expressions <code>e<sub>1</sub></code>, <code>e<sub>2</sub></code> and the result is the multiplication of the evaluation of e1 and the evaluation of e2.</li>\n\t<li>For this question, we will use a smaller subset of variable names. A variable starts with a lowercase letter, then zero or more lowercase letters or digits. Additionally, for your convenience, the names <code>&quot;add&quot;</code>, <code>&quot;let&quot;</code>, and <code>&quot;mult&quot;</code> are protected and will never be used as variable names.</li>\n\t<li>Finally, there is the concept of scope. When an expression of a variable name is evaluated, within the context of that evaluation, the innermost scope (in terms of parentheses) is checked first for the value of that variable, and then outer scopes are checked sequentially. It is guaranteed that every expression is legal. Please see the examples for more details on the scope.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> expression = &quot;(let x 2 (mult x (let x 3 y 4 (add x y))))&quot;\n<strong>Output:</strong> 14\n<strong>Explanation:</strong> 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</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> expression = &quot;(let x 3 x 2 x)&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Assignment in let statements is processed sequentially.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> expression = &quot;(let x 1 y 2 x (add x y) (add x y))&quot;\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> The first (add x y) evaluates as 3, and is assigned to x.\nThe second (add x y) evaluates as 3+2 = 5.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= expression.length &lt;= 2000</code></li>\n\t<li>There are no leading or trailing spaces in <code>expression</code>.</li>\n\t<li>All tokens are separated by a single space in <code>expression</code>.</li>\n\t<li>The answer and all intermediate calculations of that answer are guaranteed to fit in a <strong>32-bit</strong> integer.</li>\n\t<li>The expression is guaranteed to be legal and evaluate to an integer.</li>\n</ul>\n",
            "translatedTitle": "Lisp 语法解析",
            "translatedContent": "<p>给你一个类似 Lisp 语句的字符串表达式 <code>expression</code>，求出其计算结果。</p>\n\n<p>表达式语法如下所示:</p>\n\n<ul>\n\t<li>表达式可以为整数，<strong>let</strong> 表达式，<strong>add</strong> 表达式，<strong>mult</strong> 表达式，或赋值的变量。表达式的结果总是一个整数。</li>\n\t<li>(整数可以是正整数、负整数、0)</li>\n\t<li><strong>let</strong> 表达式采用&nbsp;<code>\"(let v<sub>1</sub> e<sub>1</sub> v<sub>2</sub> e<sub>2</sub> ... v<sub>n</sub> e<sub>n</sub> expr)\"</code> 的形式，其中&nbsp;<code>let</code> 总是以字符串&nbsp;<code>\"let\"</code>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&nbsp;<code>v<sub>1</sub></code>被分配为表达式&nbsp;<code>e<sub>1</sub></code>&nbsp;的值，第二个变量&nbsp;<code>v<sub>2</sub></code>&nbsp;被分配为表达式&nbsp;<code>e<sub>2</sub></code>&nbsp;的值，<strong>依次类推</strong>；最终 <code>let</code> 表达式的值为&nbsp;<code>expr</code>表达式的值。</li>\n\t<li><strong>add </strong>表达式表示为&nbsp;<code>\"(add e<sub>1</sub> e<sub>2</sub>)\"</code> ，其中&nbsp;<code>add</code> 总是以字符串&nbsp;<code>\"add\"</code> 来表示，该表达式总是包含两个表达式 <code>e<sub>1</sub></code>、<code>e<sub>2</sub></code> ，最终结果是&nbsp;<code>e<sub>1</sub></code> 表达式的值与&nbsp;<code>e<sub>2</sub></code>&nbsp;表达式的值之 <strong>和 </strong>。</li>\n\t<li><strong>mult</strong> 表达式表示为&nbsp;<code>\"(mult e<sub>1</sub> e<sub>2</sub>)\"</code>&nbsp;，其中&nbsp;<code>mult</code> 总是以字符串 <code>\"mult\"</code> 表示，该表达式总是包含两个表达式 <code>e<sub>1</sub></code>、<code>e<sub>2</sub></code>，最终结果是&nbsp;<code>e<sub>1</sub></code> 表达式的值与&nbsp;<code>e<sub>2</sub></code>&nbsp;表达式的值之<strong> 积 </strong>。</li>\n\t<li>在该题目中，变量名以小写字符开始，之后跟随 0 个或多个小写字符或数字。为了方便，<code>\"add\"</code> ，<code>\"let\"</code> ，<code>\"mult\"</code> 会被定义为 \"关键字\" ，不会用作变量名。</li>\n\t<li>最后，要说一下作用域的概念。计算变量名所对应的表达式时，在计算上下文中，首先检查最内层作用域（按括号计），然后按顺序依次检查外部作用域。测试用例中每一个表达式都是合法的。有关作用域的更多详细信息，请参阅示例。</li>\n</ul>\n&nbsp;\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>expression = \"(let x 2 (mult x (let x 3 y 4 (add x y))))\"\n<strong>输出：</strong>14\n<strong>解释：</strong>\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</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>expression = \"(let x 3 x 2 x)\"\n<strong>输出：</strong>2\n<strong>解释：</strong>let 语句中的赋值运算按顺序处理即可。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>expression = \"(let x 1 y 2 x (add x y) (add x y))\"\n<strong>输出：</strong>5\n<strong>解释：</strong>\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</pre>\n&nbsp;\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= expression.length &lt;= 2000</code></li>\n\t<li><code>exprssion</code> 中不含前导和尾随空格</li>\n\t<li><code>expressoin</code> 中的不同部分（token）之间用单个空格进行分隔</li>\n\t<li>答案和所有中间计算结果都符合 <strong>32-bit</strong> 整数范围</li>\n\t<li>测试用例中的表达式均为合法的且最终结果为整数</li>\n</ul>\n",
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