Hello Everyone,
Evaluate an expression represented by a String. The expression can contain parentheses, you can assume parentheses are well-matched. For simplicity, you can assume only binary operations allowed are +, -, *, and /. Arithmetic Expressions can be written in one of three forms:
Infix Notation: Operators are written between the operands they operate on, e.g. 3 + 4.
Prefix Notation: Operators are written before the operands, e.g + 3 4
Postfix Notation: Operators are written after operands.
Infix Expressions are harder for Computers to evaluate because of the additional work needed to decide precedence. Infix notation is how expressions are written and recognized by humans and, generally, input to programs. Given that they are harder to evaluate, they are generally converted to one of the two remaining forms. A very well known algorithm for converting an infix notation to a postfix notation is Shunting Yard Algorithm by Edgar Dijkstra. This algorithm takes as input an Infix Expression and produces a queue that has this expression converted to postfix notation. The same algorithm can be modified so that it outputs the result of the evaluation of expression instead of a queue. The trick is using two stacks instead of one, one for operands, and one for operators.
- 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. - 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. - 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.
It should be clear that this algorithm runs in linear time – each number or operator is pushed onto and popped from Stack only once.
Implementation
// 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;
}
Output:
22 212 1400 100
Time Complexity: O(n)
Space Complexity: O(n)