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aiou -solved Assighnment Question 5 -3408-data structure -spring 3015

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Solved Assighnments</h2>
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aiou -solved Assighnment -3408-daa structure </h2>
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Convert
the below mentioned infix expression to postfix </h3>
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        expression by using
the Stack.</h3>
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 a) 9+3*4/7-10</h3>
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 b) (A+B)*((D-E)+F)-G</h3>
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To watch full assignment click on this link </h3>
<h3>
    
 a) 9+3*4/7-10</h3>
</div>
#include<iostream> 
#include<cstring> 
#include<stack> 
using namespace std; 
 
// get weight of operators as per precedence 
// higher weight given to operators with higher precedence 
// for non operators, return 0 
 
int getWeight(char ch) { 
   switch (ch) { 
      case '/': 
      case '*': return 2; 
      case '+': 
      case '-': return 1; 
      default : return 0; 
   } 
} 
 
// convert infix expression to postfix using a stack 
 
void infix2postfix(char infix[], char postfix[], int size) { 
   stack<char> s; 
   int weight; 
   int i = 0; 
   int k = 0; 
   char ch; 
   // iterate over the infix expression   
   while (i < size) { 
      ch = infix[i]; 
      if (ch == '(') { 
         // simply push the opening parenthesis 
         s.push(ch); 
         i++; 
         continue; 
      } 
      if (ch == ')') { 
         // if we see a closing parenthesis, 
         // pop of all the elements and append it to 
         // the postfix expression till we encounter 
         // a opening parenthesis 
         while (!s.empty() && s.top() != '(') { 
            postfix[k++] = s.top(); 
            s.pop(); 
 
         } 
         // pop off the opening parenthesis also 
         if (!s.empty()) { 
            s.pop(); 
         } 
         i++; 
         continue; 
      } 
      weight = getWeight(ch); 
      if (weight == 0) { 
         // we saw an operand 
         // simply append it to postfix expression 
         postfix[k++] = ch; 
 
      } 
      else { 
         // we saw an operator 
         if (s.empty()) { 
            // simply push the operator onto stack if 
            // stack is empty 
            s.push(ch); 
         } 
         else { 
            // pop of all the operators from the stack and 
            // append it to the postfix expression till we 
            // see an operator with a lower precedence that 
            // the current operator 
            while (!s.empty() && s.top() != '(' && 
                  weight <= getWeight(s.top())) { 
               postfix[k++] = s.top(); 
               s.pop(); 
 
            } 
            // push the current operator onto stack 
            s.push(ch); 
         } 
      } 
      i++; 
   } 
   // pop of the remaining operators present in the stack 
   // and append it to postfix expression 
   while (!s.empty()) { 
      postfix[k++] = s.top(); 
      s.pop(); 
   } 
   postfix[k] = 0; // null terminate the postfix expression 
} 
 
// main 
int main() { 
   char infix[] = "9+3*4/7-10"; 
   int size = strlen(infix); 
   char postfix[size]; 
   infix2postfix(infix,postfix,size); 
   cout<<"\nInfix Expression :: "<<infix; 
   cout<<"\nPostfix Expression :: "<<postfix; 
   cout<<endl; 
   return 0; 
} 
 
<h3>
b) (A+B)*((D-E)+F)-G</h3>
<div style="text-align: left;">
#include<iostream> 
#include<cstring> 
#include<stack> 
using namespace std; 
 
// get weight of operators as per precedence 
// higher weight given to operators with higher precedence 
// for non operators, return 0 
 
int getWeight(char ch) { 
   switch (ch) { 
      case '/': 
      case '*': return 2; 
      case '+': 
      case '-': return 1; 
      default : return 0; 
   } 
} 
 
// convert infix expression to postfix using a stack 
 
void infix2postfix(char infix[], char postfix[], int size) { 
   stack<char> s; 
   int weight; 
   int i = 0; 
   int k = 0; 
   char ch; 
   // iterate over the infix expression   
   while (i < size) { 
      ch = infix[i]; 
      if (ch == '(') { 
         // simply push the opening parenthesis 
         s.push(ch); 
         i++; 
         continue; 
      } 
      if (ch == ')') { 
         // if we see a closing parenthesis, 
         // pop of all the elements and append it to 
         // the postfix expression till we encounter 
         // a opening parenthesis 
         while (!s.empty() && s.top() != '(') { 
            postfix[k++] = s.top(); 
            s.pop(); 
 
         } 
         // pop off the opening parenthesis also 
         if (!s.empty()) { 
            s.pop(); 
         } 
         i++; 
         continue; 
      } 
      weight = getWeight(ch); 
      if (weight == 0) { 
         // we saw an operand 
         // simply append it to postfix expression 
         postfix[k++] = ch; 
 
      } 
      else { 
         // we saw an operator 
         if (s.empty()) { 
            // simply push the operator onto stack if 
            // stack is empty 
            s.push(ch); 
         } 
         else { 
            // pop of all the operators from the stack and 
            // append it to the postfix expression till we 
            // see an operator with a lower precedence that 
            // the current operator 
            while (!s.empty() && s.top() != '(' && 
                  weight <= getWeight(s.top())) { 
               postfix[k++] = s.top(); 
               s.pop(); 
 
            } 
            // push the current operator onto stack 
            s.push(ch); 
         } 
      } 
      i++; 
   } 
   // pop of the remaining operators present in the stack 
   // and append it to postfix expression 
   while (!s.empty()) { 
      postfix[k++] = s.top(); 
      s.pop(); 
   } 
   postfix[k] = 0; // null terminate the postfix expression 
} 
 
// main 
int main() { 
   char infix[] =(A+B)*((D-E)+F)-G</div>
   int size = strlen(infix); 
   char postfix[size]; 
   infix2postfix(infix,postfix,size); 
   cout<<"\nInfix Expression :: "<<infix; 
   cout<<"\nPostfix Expression :: "<<postfix; 
   cout<<endl; 
   return 0; 
}</div>

aiou -solved Assighnment Question 5 -3408-data structure -spring 3015

Solved Assighnments AIOU aiou -solved Assighnment -3408-daa structure Convert the below mentioned infix exp...

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