String 方法按字母顺序从字符串数组中返回第一个字符串

Question

I have a problem in understanding how the following code is executed. I am seeking an example for 1/2 executions.

Code is:

public class StringArray {
  public static String getFirstString(String[] values) {
    if (values.length == 0) {
      return "";
    }

    String result = values[0];
    for (int i=1; i<values.length; i++) {
      if (result.compareTo(values[i]) > 0) {// i.e. result > values[i]
        result = values[i];
      }
    }
    return result;
  }

  public static void main(String[] args) {
    String[] nume= {"Andrei", "Andreea", "Andrea", 
                    "Marius", "Marcus", "Marcel", "Florin"};
    System.out.println(getFirstString(nume));
  }
}

Basically, is the first item processed?

First is Andrei.

1.Andrei will get into the first If? values.length should not be 7?

1.1 "value" being the reference of the parameter, should point to the array[] name which is given in the main method, right?

Therefore, Andrei will be compared to Andreea, but from here, why is Andrei bigger than Andreea? I have a hard time with the if (result.compareTo(values[i]) > 0) .

Answer 1

The key element here: understanding the "contract" of the compareTo() method.

Start by looking at the javadoc :

Compares two strings lexicographically. The comparison is based on the Unicode value of each character in the strings. The character sequence represented by this String object is compared lexicographically to the character sequence represented by the argument string. The result is a negative integer if this String object lexicographically precedes the argument string. The result is a positive integer if this String object lexicographically follows the argument string. The result is zero if the strings are equal; compareTo returns 0 exactly when the equals(Object) method would return true.

This is the definition of lexicographic ordering. If two strings are different, then either they have different characters at some index that is a valid index for both strings, or their lengths are different, or both. If they have different characters at one or more index positions, let k be the smallest such index; then the string whose character at position k has the smaller value, as determined by using the < operator, lexicographically precedes the other string. In this case, compareTo returns the difference of the two character values at position k in the two string -- that is, the value:

this.charAt(k)-anotherString.charAt(k)

If there is no index position at which they differ, then the shorter string lexicographically precedes the longer string. In this case, compareTo returns the difference of the lengths of the strings -- that is, the value:

this.length()-anotherString.length()

Answer 2

No, Your table which contains all the name (nume) goes to the first "if", so, you have 7 String, the value of your table of String is 7. The compareTo() function compare the alphabetical value, so Andrei is bigger than Andreea because "i" is bigger than "e". So, for each value of your table, you compare it with a reference value. If it's lower, you take this one in reference. You do it with all String of your list and you return the reference that is to say the lower value in order to select the last Value and you return it. So, here you should return Andrea because he has the lowest value in the list. And result should be Andrei -> Andreea -> Andrea