Question

Java程序到字母数字排序对象

请让我知道如何才能获得预期的结果

预期产量： B2 D1 D2 D14 E2

实际输出： B2 D1 D14 D2 E2

====================================

List<Name> lst = new ArrayList<>();
lst.add(new Name("D",1));
lst.add(new Name("D",14));
lst.add(new Name("D",2));
lst.add(new Name("E",2));
lst.add(new Name("B",2));

Collections.sort(lst, new Comparator<Name>() {

    @Override
    public int compare(Name n1, Name n2) {
        // TODO Auto-generated method stub
        String o1=n1.getNm()+n1.getSeatnum();
        String o2=n2.getNm()+n2.getSeatnum();


         return o1.compareTo(o2);

    }
});

for (Name name : lst) {
    System.out.println(name.getNm()+name.getSeatnum());
}

=================================

public class Name {

    private String nm;
    private int seatnum;

    public int getSeatnum() {
        return seatnum;
    }

    public void setSeatnum(int seatnum) {
        this.seatnum = seatnum;
    }

    public Name(String nm) {
        super();
        this.nm = nm;
    }

    public Name(String nm, int seatnum) {
        super();
        this.nm = nm;
        this.seatnum = seatnum;
    }

    public String getNm() {
        return nm;
    }

    public void setNm(String nm) {
        this.nm = nm;
    }

}

Answer 1

是的，但是你的compare函数需要首先检查String部分是否相等，然后对数字部分使用数字比较（目前，两者都要进行比较{{ 3}}）。所以，你可以使用像 -

这样的东西

public int compare(Name n1, Name n2) {
    int c = n1.getNm().compareTo(n2.getNm());
    if (c != 0) {
        return c;
    }
    return Integer.valueOf(n1.getSeatnum()).compareTo(n2.getSeatnum());
}

Answer 2

只需将字母与整数进行比较：

public int compare(Name n1, Name n2) {
    // TODO Auto-generated method stub
    int compare = n1.getNm().compareTo(n2.getNm());
    if (compare == 0) {
        return Integer.compare(n1.getSeatnum(), n2.getSeatnum());
    } else {
        return compare;
    }
}

Answer 3

您可以重写比较器，分两步进行：

Collections.sort(lst, new Comparator<Name>() {
    @Override
    public int compare(Name n1, Name n2) {
        // compare the name part
        int nameCompare = n1.getName().compareTo(n2.getName());
        if(nameCompare != 0)
            return nameCompare;
        // compare the number part
        return n1.getSeatnum() - n2.getSeatnum();
    }
});

如果您想了解null值，请添加：

Collections.sort(lst, new Comparator<Name>() {
    @Override
    public int compare(Name n1, Name n2) {
        // check for null Name
        if(n1 == null && n2 == null)
            return 0;
        else if(n1 == null)
            return -1;
        else if(n2 == null)
            return 1;

        // check for null in nx.getName()
        if(n1.getName() == null && n2.getName() == null)
            return 0;
        else if(n1.getName() == null)
            return -1;
        else if(n2.getName() == null)
            return 1;

        // compare the name part
        int nameCompare = n1.getName().compareTo(n2.getName());
        if(nameCompare != 0)
            return nameCompare;

        // compare the number part
        return n1.getSeatnum() - n2.getSeatnum();
    }
});

此方法会将null值放在列表的开头。如果您希望将它们放在列表的末尾，只需交换1和-1。

如果您还想要不区分大小写，请将名称比较行修改为：

        int nameCompare = n1.getName().toLowerCase().compareTo(n2.getName().toLowerCase());

Java程序到字母数字排序对象

3 个答案: