Collections.sort() 不排序单个数字

Question

This is my data set [B,A,D,C,3,10,1,2,11,14]

I want to sort it like this [1,2,3,10,11,14,A,B,C,D]

When i use following

public class ArenaModel implements Comparable<ArenaModel>{

    private String sectionName;

    @Override
    public int compareTo(ArenaModel o) {
        return sectionName.compareTo(o.sectionName);
    }

In Main class, i do following.

 Collections.sort(arenaArrayList);

It does sorting but single digit numbers do not get sort, and i get following result.

[10,11,14,1,2,3,A,B,C,D]

Answer 1

Try this:

    public class ArenaModel implements Comparable<ArenaModel>{

        private String sectionName;

        @Override
        public int compareTo(ArenaModel o) {
            try {
                Integer s1 = Integer.valueOf(sectionName);
                Integer s2 = Integer.valueOf(o.sectionName);
                return s1.compareTo(s2);
            } catch (NumberFormatException e) {
                // Not an integer
            }
            return sectionName.compareTo(o.sectionName);
        }
    }

Answer 2

Supposing that the String contains no mixed values : characters and digits, you could do that :

import static java.lang.Character.isAlphabetic;
import static java.lang.Integer.parseInt;

@Override
public int compareTo(ArenaModel o) {
    String sectionOther = o.getSectionName();
    String sectionThis = getSectionName();

    // 1) handle comparisons with any alphabetic value
    boolean isThisAlphabetic = isAlphabetic(sectionThis);
    boolean isOtherAlphabetic = isAlphabetic(sectionOther);

    // move the alphabet value at the end
    if (isThisAlphabetic && !isOtherAlphabetic){
       return 1;
    }

    // move the alphabet value at the end
    if (!isThisAlphabetic && isOtherAlphabetic){
       return -1;
    }

    // compare String values
    if (isThisAlphabetic && isOtherAlphabetic){
       return sectionThis.compareTo(sectionOther);
    }

    // 2) By eliminatation, you have two integer values
    return Integer.compare(parseInt(sectionThis), parseInt(sectionOther));
}