[Leetcode] 736. Parse Lisp Expression 解题报告

题目：

You are given a string expression representing a Lisp-like expression to return the integer value of.

The syntax for these expressions is given as follows.

• An expression is either an integer, a let-expression, an add-expression, a mult-expression, or an assigned variable. Expressions always evaluate to a single integer.

• (An integer could be positive or negative.)

• A let-expression takes the form (let v1 e1 v2 e2 ... vn en expr), where let is always the string "let", then there are 1 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.

• An add-expression takes the form (add e1 e2) where add is always the string "add", there are always two expressions e1, e2, and this expression evaluates to the addition of the evaluation of e1 and the evaluation of e2.

• A mult-expression takes the form (mult e1 e2) where mult is always the string "mult", there are always two expressions e1, e2, and this expression evaluates to the multiplication of the evaluation of e1 and the evaluation of e2.

• For the purposes of 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 "add", "let", or "mult" are protected and will never be used as variable names.

• 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 scope.

  Evaluation Examples:
  

  Input: (add 1 2)
  Output: 3
  
  Input: (mult 3 (add 2 3))
  Output: 15
  
  Input: (let x 2 (mult x 5))
  Output: 10
  
  Input: (let x 2 (mult x (let x 3 y 4 (add x y))))
  Output: 14
  Explanation: In the expression (add x y), when checking for the value of the variable x,
  we check from the innermost scope to the outermost in the context of the variable we are trying to evaluate.
  Since x = 3 is found first, the value of x is 3.
  
  Input: (let x 3 x 2 x)
  Output: 2
  Explanation: Assignment in let statements is processed sequentially.
  
  Input: (let x 1 y 2 x (add x y) (add x y))
  Output: 5
  Explanation: The first (add x y) evaluates as 3, and is assigned to x.
  The second (add x y) evaluates as 3+2 = 5.
  
  Input: (let x 2 (add (let x 3 (let x 4 x)) x))
  Output: 6
  Explanation: Even though (let x 4 x) has a deeper scope, it is outside the context
  of the final x in the add-expression.  That final x will equal 2.
  
  Input: (let a1 3 b2 (add a1 1) b2) 
  Output 4
  Explanation: Variable names can contain digits after the first character.
  
  

  Note:

• The given string expression is well formatted: There are no leading or trailing spaces, there is only a single space separating different components of the string, and no space between adjacent parentheses. The expression is guaranteed to be legal and evaluate to an integer.
• The length of expression is at most 2000. (It is also non-empty, as that would not be a legal expression.)
• The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.
  

  思路：

  采用递归思路对问题进行求解，分为如下几种情况进行处理：

  1）如果表达式是变量，那么我们就在变量表中查找到变量的名称，并且返回；如果表达式是值，则我们直接返回值。

  2）对于let而言，我们首先获得变量的名称，然后解析其对应的表达式，最后将变量的值和变量名的对应加入到表中。例如对于(let x (add 2 3) x)，我们首先得到变量名x，然后解析表达式(add 2, 3)，得到值5，然后赋值x = 5。对于最后一个x，我们在解析的过程中，直接在映射表中找到其对应的值5，并返回。

  3）对于add而言，我们首先解析第一个表达式，然后解析第二个表达式，最后将两者相加，并返回。例如对于"(add (add 2 3) (add 3 4))"而言，第一个表达式是“(add 2 3)”，第二个表达式是"(add 3 4)"，解析后我们分别得到5和7.所以最后返回12 = 5 + 7即可。

  4）对于mult而言，我们解析的方式类似于add。

  代码：

  class Solution {
  public:
      int evaluate(string expression) {
          unordered_map<string, int> myMap;
          return help(expression, myMap);
      }
  private:
      int help(string expression,unordered_map<string,int> myMap) {
          if ((expression[0] == '-') || (expression[0] >= '0' && expression[0] <= '9')) {
              return stoi(expression);
          }
          else if (expression[0] != '(') {
              return myMap[expression];
          }
          //to get rid of the first '(' and the last ')'
          string s = expression.substr(1, expression.size() - 2);
          int start = 0;
          string word = parse(s, start);
          if (word == "let") {
              while (true) {
                  string variable = parse(s, start);
                  //if there is no more expression, simply evaluate the variable
                  if (start > s.size()) {
                      return help(variable, myMap);
                  }
                  string temp = parse(s, start);
                  myMap[variable] = help(temp, myMap);                    
              }
          }
          else if (word == "add") {
              return help(parse(s, start), myMap) + help(parse(s, start), myMap);
          }
          else if (word == "mult") {
              return help(parse(s, start), myMap) * help(parse(s, start), myMap);
          } 
      }
      string parse(string &s,int &start) {    //function to seperate each expression
          int end = start + 1, temp = start, count = 1;
          if (s[start] == '(') {
              while (count != 0) {
                  if (s[end] == '(') {
                      ++count;
                  }
                  else if (s[end] == ')') {
                      --count;
                  }
                  ++end;
              }
          }
          else {
              while (end < s.size() && s[end] != ' ') {
                  ++end;
              }
          }
          start = end + 1;
          return s.substr(temp, end - temp);
      }
  };