== tests for reference equality (whether they are the same object).

.equals() tests for value equality (whether they contain the same data).

Objects.equals() checks for null before calling .equals() so you don't have to (available as of JDK7, also available in Guava).

Consequently, if you want to test whether two strings have the same value you will probably want to use Objects.equals().

// These two have the same value
new String("test").equals("test") // --> true 

// ... but they are not the same object
new String("test") == "test" // --> false 

// ... neither are these
new String("test") == new String("test") // --> false 

// ... but these are because literals are interned by 
// the compiler and thus refer to the same object
"test" == "test" // --> true 

// ... string literals are concatenated by the compiler
// and the results are interned.
"test" == "te" + "st" // --> true

// ... but you should really just call Objects.equals()
Objects.equals("test", new String("test")) // --> true
Objects.equals(null, "test") // --> false
Objects.equals(null, null) // --> true

From the Java Language Specification JLS 15.21.3. Reference Equality Operators == and !=:

While == may be used to compare references of type String, such an equality test determines whether or not the two operands refer to the same String object. The result is false if the operands are distinct String objects, even if they contain the same sequence of characters (§3.10.5, §3.10.6). The contents of two strings s and t can be tested for equality by the method invocation s.equals(t).

You almost always want to use Objects.equals(). In the rare situation where you know you're dealing with interned strings, you can use ==.

From JLS 3.10.5. String Literals:

Moreover, a string literal always refers to the same instance of class String. This is because string literals - or, more generally, strings that are the values of constant expressions (§15.28) - are "interned" so as to share unique instances, using the method String.intern.

Similar examples can also be found in JLS 3.10.5-1.

Other Methods To Consider

String.equalsIgnoreCase() value equality that ignores case. Beware, however, that this method can have unexpected results in various locale-related cases, see this question.

String.contentEquals() compares the content of the String with the content of any CharSequence (available since Java 1.5). Saves you from having to turn your StringBuffer, etc into a String before doing the equality comparison, but leaves the null checking to you.

You seem to have misunderstood what is meant by "lexicographic order".

Since c comes after b, s1 is considered bigger than s2!

Think of letters as numbers. a is 1, b is 2, c is 3, and so on. 2 comes after 1, so 2 is bigger than 1.

So comparing bb and bc is just like comparing 22 and 23. Obviously, 23 is bigger.

What you need to use is the compareTo() method of Strings.

return this.getName().compareTo(i.getName());

That should do what you want.

Usually when implementing the Comparable interface, you will just combine the results of using other Comparable members of the class.

Below is a pretty typical implementation of a compareTo() method:

class Car implements Comparable<Car> {
    int year;
    String make, model;
    public int compareTo(Car other) {
        if (!this.make.equalsIgnoreCase(other.make))
            return this.make.compareTo(other.make);
        if (!this.model.equalsIgnoreCase(other.model))
            return this.model.compareTo(other.model);
        return this.year - other.year;
    }
}