Evalúe una expresión representada por una string. La expresión puede contener paréntesis, puede suponer que los paréntesis coinciden. Para simplificar, puede suponer que solo las operaciones binarias permitidas son +, -, * y /. Las expresiones aritméticas se pueden escribir en una de tres formas:
- Notación infija: los operadores se escriben entre los operandos sobre los que operan, por ejemplo, 3 + 4.
- Notación de prefijo: los operadores se escriben antes que los operandos, por ejemplo, + 3 4
- Notación de sufijos: los operadores se escriben después de los operandos.
Las expresiones infijas son más difíciles de evaluar para las computadoras debido al trabajo adicional necesario para decidir la precedencia. La notación de infijos es cómo los humanos escriben y reconocen las expresiones y, en general, la entrada a los programas. Dado que son más difíciles de evaluar, generalmente se convierten a una de las dos formas restantes. Un algoritmo muy conocido para convertir una notación infija en una notación postfija es Shunting Yard Algorithm de Edgar Dijkstra .
Este algoritmo toma como entrada una expresión infija y produce una cola que tiene esta expresión convertida a notación de sufijo. El mismo algoritmo se puede modificar para que genere el resultado de la evaluación de la expresión en lugar de una cola. El truco está en usar dos pilas en lugar de una, una para operandos y otra para operadores.
1. While there are still tokens to be read in,
1.1 Get the next token.
1.2 If the token is:
1.2.1 A number: push it onto the value stack.
1.2.2 A variable: get its value, and push onto the value stack.
1.2.3 A left parenthesis: push it onto the operator stack.
1.2.4 A right parenthesis:
1 While the thing on top of the operator stack is not a
left parenthesis,
1 Pop the operator from the operator stack.
2 Pop the value stack twice, getting two operands.
3 Apply the operator to the operands, in the correct order.
4 Push the result onto the value stack.
2 Pop the left parenthesis from the operator stack, and discard it.
1.2.5 An operator (call it thisOp):
1 While the operator stack is not empty, and the top thing on the
operator stack has the same or greater precedence as thisOp,
1 Pop the operator from the operator stack.
2 Pop the value stack twice, getting two operands.
3 Apply the operator to the operands, in the correct order.
4 Push the result onto the value stack.
2 Push thisOp onto the operator stack.
2. While the operator stack is not empty,
1 Pop the operator from the operator stack.
2 Pop the value stack twice, getting two operands.
3 Apply the operator to the operands, in the correct order.
4 Push the result onto the value stack.
3. At this point the operator stack should be empty, and the value
stack should have only one value in it, which is the final result.
Implementación: debe quedar claro que este algoritmo se ejecuta en tiempo lineal: cada número u operador se inserta y extrae de Stack solo una vez.
C++
// CPP program to evaluate a given
// expression where tokens are
// separated by space.
#include <bits/stdc++.h>
using namespace std;
// Function to find precedence of
// operators.
int precedence(char op){
if(op == '+'||op == '-')
return 1;
if(op == '*'||op == '/')
return 2;
return 0;
}
// Function to perform arithmetic operations.
int applyOp(int a, int b, char op){
switch(op){
case '+': return a + b;
case '-': return a - b;
case '*': return a * b;
case '/': return a / b;
}
}
// Function that returns value of
// expression after evaluation.
int evaluate(string tokens){
int i;
// stack to store integer values.
stack <int> values;
// stack to store operators.
stack <char> ops;
for(i = 0; i < tokens.length(); i++){
// Current token is a whitespace,
// skip it.
if(tokens[i] == ' ')
continue;
// Current token is an opening
// brace, push it to 'ops'
else if(tokens[i] == '('){
ops.push(tokens[i]);
}
// Current token is a number, push
// it to stack for numbers.
else if(isdigit(tokens[i])){
int val = 0;
// There may be more than one
// digits in number.
while(i < tokens.length() &&
isdigit(tokens[i]))
{
val = (val*10) + (tokens[i]-'0');
i++;
}
values.push(val);
// right now the i points to
// the character next to the digit,
// since the for loop also increases
// the i, we would skip one
// token position; we need to
// decrease the value of i by 1 to
// correct the offset.
i--;
}
// Closing brace encountered, solve
// entire brace.
else if(tokens[i] == ')')
{
while(!ops.empty() && ops.top() != '(')
{
int val2 = values.top();
values.pop();
int val1 = values.top();
values.pop();
char op = ops.top();
ops.pop();
values.push(applyOp(val1, val2, op));
}
// pop opening brace.
if(!ops.empty())
ops.pop();
}
// Current token is an operator.
else
{
// While top of 'ops' has same or greater
// precedence to current token, which
// is an operator. Apply operator on top
// of 'ops' to top two elements in values stack.
while(!ops.empty() && precedence(ops.top())
>= precedence(tokens[i])){
int val2 = values.top();
values.pop();
int val1 = values.top();
values.pop();
char op = ops.top();
ops.pop();
values.push(applyOp(val1, val2, op));
}
// Push current token to 'ops'.
ops.push(tokens[i]);
}
}
// Entire expression has been parsed at this
// point, apply remaining ops to remaining
// values.
while(!ops.empty()){
int val2 = values.top();
values.pop();
int val1 = values.top();
values.pop();
char op = ops.top();
ops.pop();
values.push(applyOp(val1, val2, op));
}
// Top of 'values' contains result, return it.
return values.top();
}
int main() {
cout << evaluate("10 + 2 * 6") << "\n";
cout << evaluate("100 * 2 + 12") << "\n";
cout << evaluate("100 * ( 2 + 12 )") << "\n";
cout << evaluate("100 * ( 2 + 12 ) / 14");
return 0;
}
// This code is contributed by Nikhil jindal.
Java
/* A Java program to evaluate a
given expression where tokens
are separated by space.
*/
import java.util.Stack;
public class EvaluateString
{
public static int evaluate(String expression)
{
char[] tokens = expression.toCharArray();
// Stack for numbers: 'values'
Stack<Integer> values = new
Stack<Integer>();
// Stack for Operators: 'ops'
Stack<Character> ops = new
Stack<Character>();
for (int i = 0; i < tokens.length; i++)
{
// Current token is a
// whitespace, skip it
if (tokens[i] == ' ')
continue;
// Current token is a number,
// push it to stack for numbers
if (tokens[i] >= '0' &&
tokens[i] <= '9')
{
StringBuffer sbuf = new
StringBuffer();
// There may be more than one
// digits in number
while (i < tokens.length &&
tokens[i] >= '0' &&
tokens[i] <= '9')
sbuf.append(tokens[i++]);
values.push(Integer.parseInt(sbuf.
toString()));
// right now the i points to
// the character next to the digit,
// since the for loop also increases
// the i, we would skip one
// token position; we need to
// decrease the value of i by 1 to
// correct the offset.
i--;
}
// Current token is an opening brace,
// push it to 'ops'
else if (tokens[i] == '(')
ops.push(tokens[i]);
// Closing brace encountered,
// solve entire brace
else if (tokens[i] == ')')
{
while (ops.peek() != '(')
values.push(applyOp(ops.pop(),
values.pop(),
values.pop()));
ops.pop();
}
// Current token is an operator.
else if (tokens[i] == '+' ||
tokens[i] == '-' ||
tokens[i] == '*' ||
tokens[i] == '/')
{
// While top of 'ops' has same
// or greater precedence to current
// token, which is an operator.
// Apply operator on top of 'ops'
// to top two elements in values stack
while (!ops.empty() &&
hasPrecedence(tokens[i],
ops.peek()))
values.push(applyOp(ops.pop(),
values.pop(),
values.pop()));
// Push current token to 'ops'.
ops.push(tokens[i]);
}
}
// Entire expression has been
// parsed at this point, apply remaining
// ops to remaining values
while (!ops.empty())
values.push(applyOp(ops.pop(),
values.pop(),
values.pop()));
// Top of 'values' contains
// result, return it
return values.pop();
}
// Returns true if 'op2' has higher
// or same precedence as 'op1',
// otherwise returns false.
public static boolean hasPrecedence(
char op1, char op2)
{
if (op2 == '(' || op2 == ')')
return false;
if ((op1 == '*' || op1 == '/') &&
(op2 == '+' || op2 == '-'))
return false;
else
return true;
}
// A utility method to apply an
// operator 'op' on operands 'a'
// and 'b'. Return the result.
public static int applyOp(char op,
int b, int a)
{
switch (op)
{
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
if (b == 0)
throw new
UnsupportedOperationException(
"Cannot divide by zero");
return a / b;
}
return 0;
}
// Driver method to test above methods
public static void main(String[] args)
{
System.out.println(EvaluateString.
evaluate("10 + 2 * 6"));
System.out.println(EvaluateString.
evaluate("100 * 2 + 12"));
System.out.println(EvaluateString.
evaluate("100 * ( 2 + 12 )"));
System.out.println(EvaluateString.
evaluate("100 * ( 2 + 12 ) / 14"));
}
}
Python3
# Python3 program to evaluate a given
# expression where tokens are
# separated by space.
# Function to find precedence
# of operators.
def precedence(op):
if op == '+' or op == '-':
return 1
if op == '*' or op == '/':
return 2
return 0
# Function to perform arithmetic
# operations.
def applyOp(a, b, op):
if op == '+': return a + b
if op == '-': return a - b
if op == '*': return a * b
if op == '/': return a // b
# Function that returns value of
# expression after evaluation.
def evaluate(tokens):
# stack to store integer values.
values = []
# stack to store operators.
ops = []
i = 0
while i < len(tokens):
# Current token is a whitespace,
# skip it.
if tokens[i] == ' ':
i += 1
continue
# Current token is an opening
# brace, push it to 'ops'
elif tokens[i] == '(':
ops.append(tokens[i])
# Current token is a number, push
# it to stack for numbers.
elif tokens[i].isdigit():
val = 0
# There may be more than one
# digits in the number.
while (i < len(tokens) and
tokens[i].isdigit()):
val = (val * 10) + int(tokens[i])
i += 1
values.append(val)
# right now the i points to
# the character next to the digit,
# since the for loop also increases
# the i, we would skip one
# token position; we need to
# decrease the value of i by 1 to
# correct the offset.
i-=1
# Closing brace encountered,
# solve entire brace.
elif tokens[i] == ')':
while len(ops) != 0 and ops[-1] != '(':
val2 = values.pop()
val1 = values.pop()
op = ops.pop()
values.append(applyOp(val1, val2, op))
# pop opening brace.
ops.pop()
# Current token is an operator.
else:
# While top of 'ops' has same or
# greater precedence to current
# token, which is an operator.
# Apply operator on top of 'ops'
# to top two elements in values stack.
while (len(ops) != 0 and
precedence(ops[-1]) >=
precedence(tokens[i])):
val2 = values.pop()
val1 = values.pop()
op = ops.pop()
values.append(applyOp(val1, val2, op))
# Push current token to 'ops'.
ops.append(tokens[i])
i += 1
# Entire expression has been parsed
# at this point, apply remaining ops
# to remaining values.
while len(ops) != 0:
val2 = values.pop()
val1 = values.pop()
op = ops.pop()
values.append(applyOp(val1, val2, op))
# Top of 'values' contains result,
# return it.
return values[-1]
# Driver Code
if __name__ == "__main__":
print(evaluate("10 + 2 * 6"))
print(evaluate("100 * 2 + 12"))
print(evaluate("100 * ( 2 + 12 )"))
print(evaluate("100 * ( 2 + 12 ) / 14"))
# This code is contributed
# by Rituraj Jain
C#
/* A C# program to evaluate a given
expression where tokens
are separated by space.
*/
using System;
using System.Collections.Generic;
using System.Text;
public class EvaluateString
{
public static int evaluate(string expression)
{
char[] tokens = expression.ToCharArray();
// Stack for numbers: 'values'
Stack<int> values = new Stack<int>();
// Stack for Operators: 'ops'
Stack<char> ops = new Stack<char>();
for (int i = 0; i < tokens.Length; i++)
{
// Current token is a whitespace, skip it
if (tokens[i] == ' ')
{
continue;
}
// Current token is a number,
// push it to stack for numbers
if (tokens[i] >= '0' && tokens[i] <= '9')
{
StringBuilder sbuf = new StringBuilder();
// There may be more than
// one digits in number
while (i < tokens.Length &&
tokens[i] >= '0' &&
tokens[i] <= '9')
{
sbuf.Append(tokens[i++]);
}
values.Push(int.Parse(sbuf.ToString()));
// Right now the i points to
// the character next to the digit,
// since the for loop also increases
// the i, we would skip one
// token position; we need to
// decrease the value of i by 1 to
// correct the offset.
i--;
}
// Current token is an opening
// brace, push it to 'ops'
else if (tokens[i] == '(')
{
ops.Push(tokens[i]);
}
// Closing brace encountered,
// solve entire brace
else if (tokens[i] == ')')
{
while (ops.Peek() != '(')
{
values.Push(applyOp(ops.Pop(),
values.Pop(),
values.Pop()));
}
ops.Pop();
}
// Current token is an operator.
else if (tokens[i] == '+' ||
tokens[i] == '-' ||
tokens[i] == '*' ||
tokens[i] == '/')
{
// While top of 'ops' has same
// or greater precedence to current
// token, which is an operator.
// Apply operator on top of 'ops'
// to top two elements in values stack
while (ops.Count > 0 &&
hasPrecedence(tokens[i],
ops.Peek()))
{
values.Push(applyOp(ops.Pop(),
values.Pop(),
values.Pop()));
}
// Push current token to 'ops'.
ops.Push(tokens[i]);
}
}
// Entire expression has been
// parsed at this point, apply remaining
// ops to remaining values
while (ops.Count > 0)
{
values.Push(applyOp(ops.Pop(),
values.Pop(),
values.Pop()));
}
// Top of 'values' contains
// result, return it
return values.Pop();
}
// Returns true if 'op2' has
// higher or same precedence as 'op1',
// otherwise returns false.
public static bool hasPrecedence(char op1,
char op2)
{
if (op2 == '(' || op2 == ')')
{
return false;
}
if ((op1 == '*' || op1 == '/') &&
(op2 == '+' || op2 == '-'))
{
return false;
}
else
{
return true;
}
}
// A utility method to apply an
// operator 'op' on operands 'a'
// and 'b'. Return the result.
public static int applyOp(char op,
int b, int a)
{
switch (op)
{
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
if (b == 0)
{
throw new
System.NotSupportedException(
"Cannot divide by zero");
}
return a / b;
}
return 0;
}
// Driver method to test above methods
public static void Main(string[] args)
{
Console.WriteLine(EvaluateString.
evaluate("10 + 2 * 6"));
Console.WriteLine(EvaluateString.
evaluate("100 * 2 + 12"));
Console.WriteLine(EvaluateString.
evaluate("100 * ( 2 + 12 )"));
Console.WriteLine(EvaluateString.
evaluate("100 * ( 2 + 12 ) / 14"));
}
}
// This code is contributed by Shrikant13
Javascript
<script>
/* A Javascript program to evaluate a given
expression where tokens
are separated by space.
*/
function evaluate(expression)
{
let tokens = expression.split('');
// Stack for numbers: 'values'
let values = [];
// Stack for Operators: 'ops'
let ops = [];
for (let i = 0; i < tokens.length; i++)
{
// Current token is a whitespace, skip it
if (tokens[i] == ' ')
{
continue;
}
// Current token is a number,
// push it to stack for numbers
if (tokens[i] >= '0' && tokens[i] <= '9')
{
let sbuf = "";
// There may be more than
// one digits in number
while (i < tokens.length &&
tokens[i] >= '0' &&
tokens[i] <= '9')
{
sbuf = sbuf + tokens[i++];
}
values.push(parseInt(sbuf, 10));
// Right now the i points to
// the character next to the digit,
// since the for loop also increases
// the i, we would skip one
// token position; we need to
// decrease the value of i by 1 to
// correct the offset.
i--;
}
// Current token is an opening
// brace, push it to 'ops'
else if (tokens[i] == '(')
{
ops.push(tokens[i]);
}
// Closing brace encountered,
// solve entire brace
else if (tokens[i] == ')')
{
while (ops[ops.length - 1] != '(')
{
values.push(applyOp(ops.pop(),
values.pop(),
values.pop()));
}
ops.pop();
}
// Current token is an operator.
else if (tokens[i] == '+' ||
tokens[i] == '-' ||
tokens[i] == '*' ||
tokens[i] == '/')
{
// While top of 'ops' has same
// or greater precedence to current
// token, which is an operator.
// Apply operator on top of 'ops'
// to top two elements in values stack
while (ops.length > 0 &&
hasPrecedence(tokens[i],
ops[ops.length - 1]))
{
values.push(applyOp(ops.pop(),
values.pop(),
values.pop()));
}
// Push current token to 'ops'.
ops.push(tokens[i]);
}
}
// Entire expression has been
// parsed at this point, apply remaining
// ops to remaining values
while (ops.length > 0)
{
values.push(applyOp(ops.pop(),
values.pop(),
values.pop()));
}
// Top of 'values' contains
// result, return it
return values.pop();
}
// Returns true if 'op2' has
// higher or same precedence as 'op1',
// otherwise returns false.
function hasPrecedence(op1, op2)
{
if (op2 == '(' || op2 == ')')
{
return false;
}
if ((op1 == '*' || op1 == '/') &&
(op2 == '+' || op2 == '-'))
{
return false;
}
else
{
return true;
}
}
// A utility method to apply an
// operator 'op' on operands 'a'
// and 'b'. Return the result.
function applyOp(op, b, a)
{
switch (op)
{
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
if (b == 0)
{
document.write("Cannot divide by zero");
}
return parseInt(a / b, 10);
}
return 0;
}
document.write(evaluate("10 + 2 * 6") + "</br>");
document.write(evaluate("100 * 2 + 12") + "</br>");
document.write(evaluate("100 * ( 2 + 12 )") + "</br>");
document.write(evaluate("100 * ( 2 + 12 ) / 14") + "</br>");
// This code is contributed by decode2207.
</script>
Producción
22 212 1400 100
Complejidad de tiempo: O(n)
Complejidad de espacio: O(n)
Vea esto para una ejecución de muestra con más casos de prueba.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA