1.

Write a function to convert an infix expression to postfix expression

Answer»


  • Input: a+b*(c^d)


  • Output: abcd^*+

int prec(char c)
{
if (c == '^')
return 3;
else if (c == '/' || c == '*')
return 2;
else if (c == '+' || c == '-')
return 1;
else
return -1;
}
public:
// Function to convert an infix expression to a postfix expression.
string infixToPostfix(string s) {
stack<char> st; // For stack operations, we are using C++ built in stack
string result;

for (int i = 0; i < s.length(); i++) {
char c = s[i];

// If the scanned character is
// an operand, add it to the output string.
if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')
|| (c >= '0' && c <= '9'))
result += c;

// If the scanned character is an
// '(', push it to the stack.
else if (c == '(')
st.push('(');

// If the scanned character is an ')',
// pop and to output string from the stack
// until an '(' is encountered.
else if (c == ')') {
while (st.top() != '(') {
result += st.top();
st.pop();
}
st.pop();
}

// If an operator is scanned
else {
while (!st.empty()
&& prec(s[i]) <= prec(st.top())) {
if (c == '^' && st.top() == '^')
break;
else {
result += st.top();
st.pop();
}
}
st.push(c);
}
}

// Pop all the remaining elements from the stack
while (!st.empty()) {
result += st.top();
st.pop();
}

return result;
}


  • Time Complexity: O(n)


  • Space Complexity: O(n)




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