Sorting algorithm without using compareTo

Question

I'm working on a homework assignment that requires me to compare two strings and determine if they're in alphabetical order.

I plan to write a method that will take two strings as arguments, (String a, String b) and return either 1, 0, or -1 (so, an int) signalling whether a > b, a < b, or otherwise (the 0 case).

For example, comparing ("boogie", "orange") would return a -1. since, boogie < orange.

My code so far is

public static int compare(String a, String b) {
    for (int i = 0; i < a.length(); i++) {
        for (int j = 0; j < b.length(); j++) {
            char cha = a.charAt(i);
            char chb = b.charAt(j);
            if (cha < chb) {
                return -1;
            } else if (cha > chb) {
                return 1;
            }
        }
        return 0;
    }
}

However, I am encountering numerous errors and cannot find fixes for the bugs. I'm also having difficulty finding a code for measuring if one word is longer than another (which affects alphabetical order) Can someone help me debug the code and point me in the right direction?

Many thanks in advance.

If you're getting errors, always post them. Don't make us sit here and guess! — tnw
– tnw, Commented Mar 13, 2015 at 19:41
"I'm also having difficulty finding a code for measuring if one word is longer than another (which affects alphabetical order)" ...length()? — Tetramputechture
– Tetramputechture, Commented Mar 13, 2015 at 19:42
When you say alphabetical order between the two words what do you mean: the first letter of each word? All letters compared to each other? Right now you are only comparing first characters in each word because as soon as you have a match cha > chb or cha < chb it returns. It will also return 0 right now if both words are identical. — Code Whisperer
– Code Whisperer, Commented Mar 13, 2015 at 19:43

Eran · Accepted Answer · 2015-03-13 19:55:52Z

2

You don't need a nestd loop, since you don't want to compare every character of one String to every character of the other String.

You only need a single loop:

public static int compare(String a, String b)
{
    int len = Math.min (a.length(),b.length());
    for (int i = 0; i<len; i++) {
        char cha = a.charAt(i);
        char chb = b.charAt(i);
        if (cha < chb) {
            return -1;
        } else if (cha > chb) {
            return 1; 
        }
    }
    if (a.length() < b.length())
        return -1;
    else if (a.length() > b.length())
        return 1;
    else
        return 0;
}

As for handling Strings of different lengths, if you find that the shorter of the 2 Strings is equal to the prefix of the longer String, you return -1 if a is the shorter String and 1 if b is shorter (since the shorter String should come before the longer one).

edited Mar 13, 2015 at 19:55

answered Mar 13, 2015 at 19:40

Eran

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13 Comments

Apurva Over a year ago

can characters be compared by < or >?

Eran Over a year ago

@Apurva Characters have integer value between 0 and 2^16-1, so they can be compared with < or >.

Piyush Vishwakarma Over a year ago

Yes, in java you can compare character with <,>,<=,>=,== etc. just like Integers.

Eran Over a year ago

@CodeWhisperer Based on lexicographical order abcde < abcdexx (regardless of what characters appear in place of xx). The longer word would always be larger than the shorter word if the longer word starts with the shorter word.

Eran Over a year ago

@Aify No, it wouldn't, since equals also iterates over the characters of the two Strings (assuming they have equal length), so you'll be running some of the logic twice.

|

Vincent C · Accepted Answer · 2015-03-13 20:37:02Z

-1

You can compare two char by using '-' operator rather than the '>'. For example below.

public static int compare(String a, String b) {
   return  a.charAt(0) - b.charAt(0);
}

In your case, something like this.

public static int compare(char cha, char chb) {
    if (cha-chb < 0) {
         return -1;
     } else if(chb - cha > 0){
         return 1;
     } else if(chb - cha == 0){
         return 0;
     }
    return 0;
}

answered Mar 13, 2015 at 20:37

Vincent C

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