Question

I am trying to write a method that uses recursion to compare the strings str1 and str2 and determine which of them comes first alphabetically (i.e., according to the ordering used for words in a dictionary).

If str1 comes first alphabetically, the method should return int 1.

If str2 comes first alphabetically, the method should return the int 2.

If the two strings are the same, the method should return the int 0.

I know that there is a compareTo method in the Java API but i would like to know how to do this without this

This is what i have so far but i m not entirely sure how to proceeded

 } if (str1.length().equals(str2.length()))

         return 0;
    } else {
        (str.substring(1, str.length()));

Any ideas would be greatly appreciated

Answer 1

Make a method int recursiveCompare(String string1, String string2, int index). Initially call it with index = 0. Compare string1.charAt(index) and string2.charAt(index), and if they re different, return 1 or 2. If they re the same, return recursiveCompare(string1, string2, index + 1).

Of course, you ll have to check the lengths of string1 and string2 before calling charAt(index). If they both reach the end at the same time, they re equal, so return 0. Otherwise, return the number of the one that has ended.

And yeah, recursion is pretty much the worst way to do this, LOL.

Answer 2

No recursion required... (Unless this is specifically required in the homework(?) assignement...)

As this looks a lot like homework, I ll just give a few hints

Use a integer variable, say i, to index from 0 to the length of shorter string. As long as str1[i] == str2[i], and the last index value hasn t been reached, increment i. If you do reach the last possible value for the index, then the shorter string comes first (or they are deemed equal if same length...)

Otherwise, compare this first character that differs, and decide which string is first accordingly... Could be as simple as:
return (str1[i] < str2[i]);

If recursion you must... (and it was readily said in other comments, this kind of problem is truly not a logical/valid candidate for recursion...)

The idea is to have a function with this kind of interface:

int  RecursCompare(string str1, string str2, int i)

and which calls itself, passing the same values for str1 an str2 and passing the next value for i (i+1), as long as str1[i] == str2[i] AND neither str1 or str2 is at its end. When this condidtion becomes false, recursion ends, and instead the function returns the appropriate value to signify tha Str1 is alphabetically before or after Str2.

Answer 3

#include<stdio.h>
main()
{
char str1[100],str2[100];
int i=0,k=0;
puts("Enter string 1");
gets(str1);
puts("Enter string 2");
gets(str2);
i=comp(str1,str2,0);
printf(" count is %d %d ",i,strlen(str1));
(strlen(str1)==strlen(str2)) ?( (strlen(str1)==i) ? printf("Both are equal"):printf("Both are Not equal")):printf("Both are Not equal");

}
int comp(char s1[], char s2[],int i)
{
printf(" %c %c",s1[i],s2[i]);
int sum=0,count =1;
if((s1[i] != )|| (s2[i]!= ))
{
if (s1[i] == s2[i])
{
return (count += comp(s1,s2,++i));
}
else
{
return 0;
}
}
else
{
return 0;
}
return count;

}

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