Problem
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, let expression, add expression, 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 one or more pairs of alternating variables and expressions, meaning that the first variablev1is assigned the value of the expressione1, the second variablev2is assigned the value of the expressione2, and so on sequentially; and then the value of this let expression is the value of the expressionexpr.An add expression takes the form
"(add e1 e2)"where add is always the string"add", there are always two expressionse1,e2and the result is the addition of the evaluation ofe1and the evaluation ofe2.A mult expression takes the form
"(mult e1 e2)"where mult is always the string"mult", there are always two expressionse1,e2and the result is the multiplication of the evaluation of e1 and the evaluation of e2.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
"add","let", and"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 the scope.
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 <= 2000There are no leading or trailing spaces in
expression.All tokens are separated by a single space in
expression.The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.
The expression is guaranteed to be legal and evaluate to an integer.
Solution (Java)
class Solution {
static class Exp {
Deque<Exp> exps;
String op;
Exp parent;
public Exp(Exp from) {
this.exps = new LinkedList<>();
this.parent = from;
}
private int evaluate(Map<String, Integer> vars) {
if (op.equalsIgnoreCase("add")) {
return exps.pop().evaluate(vars) + exps.pop().evaluate(vars);
} else if (op.equalsIgnoreCase("mult")) {
return exps.pop().evaluate(vars) * exps.pop().evaluate(vars);
} else if (op.equalsIgnoreCase("let")) {
Map<String, Integer> nextVars = new HashMap<>(vars);
while (exps.size() > 1) {
String varName = exps.pop().op;
int val = exps.pop().evaluate(nextVars);
nextVars.put(varName, val);
}
return exps.pop().evaluate(nextVars);
} else {
if (Character.isLetter(op.charAt(0))) {
return vars.get(op);
} else {
return Integer.parseInt(op);
}
}
}
}
private Exp buildTree(String exp) {
Exp root = new Exp(null);
Exp cur = root;
int n = exp.length() - 1;
while (n >= 0) {
char c = exp.charAt(n);
if (c == ')') {
Exp next = new Exp(cur);
cur.exps.push(next);
cur = next;
} else if (c == '(') {
cur.op = cur.exps.pop().op;
cur = cur.parent;
} else if (c != ' ') {
int pre = n;
while (pre >= 0 && exp.charAt(pre) != '(' && exp.charAt(pre) != ' ') {
pre--;
}
Exp next = new Exp(cur);
next.op = exp.substring(pre + 1, n + 1);
cur.exps.push(next);
n = pre + 1;
}
n--;
}
return root.exps.pop();
}
public int evaluate(String exp) {
return buildTree(exp).evaluate(new HashMap<>());
}
}
Solution (C++)
class Solution {
public:
int evaluate(string expression) {
scopes_.clear();
int pos = 0;
return eval(expression, pos);
}
private:
int eval(const string& s, int& pos) {
scopes_.push_front(unordered_map<string, int>());
int value = 0; // The return value of current expr
if (s[pos] == '(') ++pos;
// command, variable or number
const string token = getToken(s, pos);
if (token == "add") {
int v1 = eval(s, ++pos);
int v2 = eval(s, ++pos);
value = v1 + v2;
} else if (token == "mult") {
int v1 = eval(s, ++pos);
int v2 = eval(s, ++pos);
value = v1 * v2;
} else if (token == "let") {
string var;
// expecting " var1 exp1 var2 exp2 ... last_expr)"
while (s[pos] != ')') {
++pos;
// Must be last_expr
if (s[pos] == '(') {
value = eval(s, ++pos);
break;
}
// Get a token, could be "x" or "-12" for last_expr
var = getToken(s, pos);
// End of let, var is last_expr
if (s[pos] == ')') {
if (isalpha(var[0]))
value = getValue(var);
else
value = stoi(var);
break;
}
// x -12 -> set x to -12 and store in the current scope and take it as the current return value
value = scopes_.front()[var] = eval(s, ++pos);
}
} else if (isalpha(token[0])) {
value = getValue(token); // symbol
} else {
value = std::stoi(token); // number
}
if (s[pos] == ')') ++pos;
scopes_.pop_front();
return value;
}
int getValue(const string& symbol) {
for (const auto& scope : scopes_)
if (scope.count(symbol)) return scope.at(symbol);
return 0;
}
// Get a token from current pos.
// "let x" -> "let"
// "-12 (add x y)" -> "-12"
string getToken(const string& s, int& pos) {
string token;
while (pos < s.length()) {
if (s[pos] == ')' || s[pos] == ' ') break;
token += s[pos++];
}
return token;
}
deque<unordered_map<string, int>> scopes_;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).