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 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
.(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
.(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
.
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:
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.expression
is at most 2000. (It is also non-empty, as that would not be a legal expression.)
要点:剥洋葱
class Solution { public int evaluate(String expression) { return eval(expression, new HashMap<String, Integer>()); } private int eval(String exp, Map<String, Integer> parent) { if (exp.charAt(0) != '(') { if (Character.isDigit(exp.charAt(0)) || exp.charAt(0) == '-') { return Integer.parseInt(exp); } return parent.get(exp); } Map<String, Integer> local = new HashMap<>(parent); List<String> tokens; if (exp.startsWith("(mult")) { tokens = parse(exp.substring(6, exp.length() - 1)); return eval(tokens.get(0), local) * eval(tokens.get(1), local); }else if (exp.startsWith("(add")) { tokens = parse(exp.substring(5, exp.length() - 1)); return eval(tokens.get(0), local) + eval(tokens.get(1), local); }else { tokens = parse(exp.substring(5, exp.length() - 1)); for (int i = 0; i < tokens.size() - 1; i += 2) { local.put(tokens.get(i), eval(tokens.get(i + 1), local)); } return eval(tokens.get(tokens.size() - 1), local); } } private List<String> parse(String s) { List<String> rt = new ArrayList<>(); int count = 0; String sb = ""; for (int i = 0; i < s.length(); i++) { if (s.charAt(i) == '(') count++; if (s.charAt(i) == ')') count--; if (count == 0 && s.charAt(i) == ' ') { rt.add(sb); sb = ""; }else { sb += s.charAt(i); } } if (sb.length() > 0) rt.add(sb); return rt; } }
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