public class Solution extends Object
736 - Parse Lisp Expression.
Hard
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.
“(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.Example 1:
Input: expression = “(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.
Example 2:
Input: expression = “(let x 3 x 2 x)”
Output: 2
Explanation: Assignment in let statements is processed sequentially.
Example 3:
Input: expression = “(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.
Constraints:
1 <= expression.length <= 2000
expression
.expression
.Constructor and Description |
---|
Solution() |
public int evaluate(String exp)
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