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.)struct Solution;
use std::collections::HashMap;
use std::iter::Peekable;
use std::str::Chars;
use std::vec::IntoIter;
#[derive(Debug, Clone, Copy, Eq, PartialEq)]
enum Tok {
Var(usize),
Num(i32),
Op(char),
}
use Tok::*;
impl Solution {
fn evaluate(expression: String) -> i32 {
let mut it = expression.chars().peekable();
let mut symbols: HashMap<String, usize> = HashMap::new();
let tokens = Self::parse_tokens(&mut it, &mut symbols);
let mut it = tokens.into_iter().peekable();
let k = symbols.len();
let mut stacks = vec![vec![]; k];
Self::eval_expr(&mut it, &mut stacks).unwrap()
}
fn eval_expr(it: &mut Peekable<IntoIter<Tok>>, stacks: &mut Vec<Vec<i32>>) -> Option<i32> {
let tok = it.next().unwrap();
match tok {
Op('(') => {
let res = Self::eval_complex_expr(it, stacks).unwrap();
let close = it.next().unwrap();
if close != Op(')') {
panic!();
}
Some(res)
}
Var(id) => Some(stacks[id].last().copied().unwrap()),
Num(x) => Some(x),
_ => {
panic!()
}
}
}
fn eval_complex_expr(
it: &mut Peekable<IntoIter<Tok>>,
stacks: &mut Vec<Vec<i32>>,
) -> Option<i32> {
let tok = it.next().unwrap();
match tok {
Op('=') => {
let mut locals = vec![];
let mut ambiguous: Option<i32> = None;
while let Some(Var(_)) = it.peek() {
if let Some(Var(id)) = it.next() {
if let Some(Op(')')) = it.peek() {
let val = stacks[id].last().copied().unwrap();
ambiguous = Some(val);
break;
} else {
let expr = Self::eval_expr(it, stacks).unwrap();
stacks[id].push(expr);
locals.push(id);
}
} else {
panic!();
}
}
let res = if let Some(val) = ambiguous {
val
} else {
Self::eval_expr(it, stacks).unwrap()
};
for id in locals {
stacks[id].pop().unwrap();
}
Some(res)
}
Op('+') => {
Some(Self::eval_expr(it, stacks).unwrap() + Self::eval_expr(it, stacks).unwrap())
}
Op('*') => {
Some(Self::eval_expr(it, stacks).unwrap() * Self::eval_expr(it, stacks).unwrap())
}
_ => panic!(),
}
}
fn parse_tokens(it: &mut Peekable<Chars>, symbols: &mut HashMap<String, usize>) -> Vec<Tok> {
let mut res = vec![];
while let Some(c) = it.next() {
match c {
'(' | ')' => res.push(Tok::Op(c)),
'-' => {
let mut x = 0;
while let Some(next_c) = it.peek() {
if next_c.is_numeric() {
x *= 10;
x += (it.next().unwrap() as u8 - b'0') as i32;
} else {
break;
}
}
res.push(Tok::Num(-x));
}
'0'..='9' => {
let mut x = (c as u8 - b'0') as i32;
while let Some(next_c) = it.peek() {
if next_c.is_numeric() {
x *= 10;
x += (it.next().unwrap() as u8 - b'0') as i32;
} else {
break;
}
}
res.push(Tok::Num(x));
}
'a'..='z' => {
let mut s = "".to_string();
s.push(c);
while let Some(next_c) = it.peek() {
if next_c.is_alphanumeric() {
s.push(it.next().unwrap());
} else {
break;
}
}
match s.as_str() {
"let" => res.push(Tok::Op('=')),
"mult" => res.push(Tok::Op('*')),
"add" => res.push(Tok::Op('+')),
_ => {
let size = symbols.len();
let id = *symbols.entry(s).or_insert(size);
res.push(Tok::Var(id));
}
}
}
' ' => {}
_ => panic!(),
}
}
res
}
}
#[test]
fn test() {
let expression = "(add 1 2)".to_string();
let res = 3;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(mult 3 (add 2 3))".to_string();
let res = 15;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(let x 2 (mult x 5))".to_string();
let res = 10;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(let x 2 (mult x (let x 3 y 4 (add x y))))".to_string();
let res = 14;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(let x 3 x 2 x)".to_string();
let res = 2;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(let x 1 y 2 x (add x y) (add x y))".to_string();
let res = 5;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(let x 2 (add (let x 3 (let x 4 x)) x))".to_string();
let res = 6;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(let a1 3 b2 (add a1 1) b2)".to_string();
let res = 4;
assert_eq!(Solution::evaluate(expression), res);
let expression = "(let x 7 -12)".to_string();
let res = -12;
assert_eq!(Solution::evaluate(expression), res);
}