“I bought this guide a few days ago to prepare for my interview with Oracle. Many of the questions they asked me were from this guide. I found this book absolutely great!”
by the end of the process, the address os s would have changed and you won’t get the correct answer. the solution would be to use a char* variable. ie.. add the following sentense before the while loop char *res=s; and return res
The function “Reverse” is called with the start address of the string to be reversed and number of characters to be reversed. This is useful because this function can also be used to reverse a portion of a string instead of an entire string, and such a function can also be used in “reversing words in a sentence”, another famous question from Microsoft.
//string to hold letters
char cString[8] = “reverse”;
//iterators for the reverse
int iBegin = 0;
//this int has to hold the number of characters in the word.
//use strlen or something, there are loads
//out there, i’m just trying to keep this simple
int iEnd = 6;
//the numbers should not cross becuase your swapping
//the front letters with the back, if the numbers meet
//then you have an odd number of characters and the
//middle one should stay the same
while((iBegin > iEnd;
(10 votes, avg: 3.4)
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strrev(string_to_reverse);
I actually gave the answer above to a MS interviewer. He was not impressed.
void rev(char str[], int len)
{
if (len>0)
{
printf(”%c”,str[len-1]);
rev(str,len-1);
}
}
void strrev (char *s)
{
char *t = s + strlen(s) - 1;
char ch = ‘ ’;
while (s < t) {
ch = *s;
*s = *t;
*t = ch;
s++;
t–;
}
}
by the end of the process, the address os s would have changed and you won’t get the correct answer. the solution would be to use a char* variable.
ie.. add the following sentense before the while loop
char *res=s;
and return res
this method uses only 2 bytes of memory
revstr(char* str)
{
char* a = str;
char* b = str+ (strlen(str) - 1);
while(a < b)
{
*a = *a + *b;
*b = *a - *b;
*a = *a - *b;
++a;
–b;
}
}
well the above solution is a goog one.If you are thinking of reversing a string using recursion i think this is the solution.
reverse( char *s)
{
static int i=0;
while( *s != ‘/0)
{
if (i==0)
{
reverse(*s);
i++;
}
else
reverse(++s*);
}/* end of while loop*/
putchar(*s);
}
revstr(char* s)
{
int len=strlen(s);
int i;
char t;
for (i=0; i<len/2; i++) {
c = s[i];
s[i] = s[len - i];
s[len - i] = c;
}
}
JV’s idea is wrong.
The address is in a local variable of a callee method.
So it doesn’t change the address in a caller.
char* strRev( char* pStr )
{
char *pStart = pStr;
char *pEnd = pStr;
char c;
while( *pStr++ );
pStr -= 2;
while ( pEnd < pStr )
{
c = *pEnd;
*pEnd++ = *pStr;
*pStr — = c;
}
return pStart;
}
str_rev(char s[])
{
int len = strlen(s);
int i;
char temp;
for(i=0; i<len/2; i++)
{
temp = s[i];
s[i] = s[(len-1) - i];
s[(len-1) - i] = temp;
}
}
void main()
{
char ch;
if( (ch = getchar()) != ‘n’ )
main();
printf(”%c”, ch);
}
The function “Reverse” is called with the start address of the string to be reversed and number of characters to be reversed. This is useful because this function can also be used to reverse a portion of a string instead of an entire string, and such a function can also be used in “reversing words in a sentence”, another famous question from Microsoft.
void Reverse(char * S, int N)
{
char * Start=S;
char * End=&S[N-1];
char Temp;
while (Start<End)
{
Temp=*Start;
*Start=*End;
*End=Temp;
Start++;
End–;
}
}
To reverse a string make a call this way:
int main()
{
char Str[100]={”This is a sample string”};
// Reverse entire string
Reverse(Str, strlen(Str));
// To reverse the first word only
Reverse(Str, 4);
// To reverse the fourth word only
Reverse(Str+10, 6);
}
void str_rev(char *s)
{
char *p;
int i;
for (i=0;*(s+i) != ‘ ’ ;i++ ){}
p = (s+i);
for (;i >= 0 ;i– )
{
putchar(*p);
p–;
}
}
This is not string reverse, but it prints the input string in the reverse order.
void rev( char *s)
{
if (*s) rev(s+1), cout << *s ;
}
Here is the recursive version of rev method.
#include <iostream.h>
char* rev( char *s, char* t = 0)
{
if (!t) t = s;
if (*s)
{
char* r = rev(s+1, t);
if (r < s) *r ^= *s ^= *r ^= *s;
return r+1;
}
return t;
}
void main()
{
char even_str[] = “Hell”;
char odd_str[] = “Hello”;
char null_str[] = “”;
rev(even_str); cout << even_str << ‘n’;
rev(odd_str); cout << odd_str << ‘n’;
rev(null_str); cout << null_str << ‘n’;
}
Here’s a C# solution:
using System;
namespace Reverser {
public class StringReverse {
static String reverse(String s) {
String result = null;
for(int i = 0; i <= s.Length; i++) {
if (i > 0)
result += s[s.Length - i];
}
return result;
}
public static void Main(String []args) {
String test = “Reverse this String!”;
Console.WriteLine(”Output: ” + reverse(test));
}
}
}
Using C#, you should rather use a StringBuilder with a fixed size of the original string. It’s far better in termes of perf and memory
Here is a Java Solution
public String reverseString(String str, int n)
{ String temp = “”;
if( n >= str.length() -1)
return temp + str.charAt(str.length() -1);// basis
return reverseString(str, n + 1) + str.charAt(n); // recursion
}
Recursive solution:
char * StrRev(char * s)
{
char t;
if(strlen(s) <=1)
return s;
t = *s;
strcpy(s,StrRev(s+1));
*(s+strlen(s)) = t;
return s;
}
Yet another…
char * revStr(char *s){
int len = 0;
char *rr = NULL;
while(*s){len++;s++;}
rr = (char*)malloc((len+1)*sizeof(char));
rr[len] = ”;
while(*rr){
*rr = *–s;
rr++;
}
return rr - len;
}
while(index_low
//string to hold letters
char cString[8] = “reverse”;
//iterators for the reverse
int iBegin = 0;
//this int has to hold the number of characters in the word.
//use strlen or something, there are loads
//out there, i’m just trying to keep this simple
int iEnd = 6;
//the numbers should not cross becuase your swapping
//the front letters with the back, if the numbers meet
//then you have an odd number of characters and the
//middle one should stay the same
while((iBegin > iEnd;
function string_reverse(char str[n])
{
i = 0;
j = n-1;
while(i less than j)
{
swap(str[i++], str[j–]);
}
}
Copy from somewhere,
include “string”
string strRev( const string & s ) {
return (s.size()==1 ? s : strRev(s.substr(1)) + *s.begin());
};
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