Answer / sandeep

node * reverse(node * list)
{
node *p, *q, *r;
p = list;
q = p->next;
while(q->next != NULL)
{
q = p->next;
r = q->next;
p->next = r;
q->next = p;
p = p->next;
}
q->next = p;
p->next = NULL;
return q`;
}

Answer / mahesh

refer to the page number 456 in ds by sahani

Answer / hasan ali mirza

list reverse(list L)
{
stack S;
position Lpos = first(L);
while(Lpos->element != NULL)
{
push(Lpos->element, S);
Lpos = Lpos->next;
}
makeEmpty(L);
Lpos = first(L);
while(isempty(S) == FALSE)
{
insert(pop(S), Lpos);
Lpos = Lpos->next;
}
return L;
}

Answer / hasan ali mirza

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int preceed(char c);
void push(char c);
char pop();
char stk[30];
int tos=-1;
int main()
{
int i,j=0,n,u,v;
char infix[30],postfix[30],c;
printf("Enter the infix expression:");
scanf("%s",&infix);
n=strlen(infix);
for(i=0;i<n;i++)
{
if((infix[i]>='a'&&infix[i]<='z')||(infix[i]>='A'&&infix[i]<='Z'))
{
postfix[j]=infix[i];
j++;
}
else if(infix[i]=='^'||infix[i]=='*'||infix[i]=='/'||infix[i]=='+'||infix[i]=='-')
{
u=preceed(stk[tos]);
v=preceed(infix[i]);
while(u>=v&&stk[tos]!='(')
{
postfix[j]=pop();
j++;
u=preceed(stk[tos]);
}
push(infix[i]);
}
else if(infix[i]=='(')
{
push(infix[i]);
}
else if(infix[i]==')')
{
c=pop();
while(c!='(')
{
postfix[j]=c;
j++;
c=pop();
}
}
else
{
printf("Equation has error\n");
exit(0);
}
}
while(tos!=-1)
{
postfix[j]=pop();
j++;
}
postfix[j]=='\0';
printf("The equation in postfix:%s",postfix);
}
void push(char c)
{
tos++;
stk[tos]=c;
}
char pop()
{
char val;
val=stk[tos];
tos--;
return(val);
}

int preceed(char c)
{
int v;
switch(c)
{
case '^':v=3;
break;
case '*':
case '/':v=2;
break;
case '+':
case '-':v=1;
break;
default:v=0;
break;
}
return(v);
}

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