java - Java排序算法字符串与数字

Question

我有一个字符串列表。示例_1，示例_2，示例_3 ...示例_99 在这里排序的最佳算法是什么？

可以使用 Collat​​or 吗？这是我目前的程序，但我想可能有更好的方法：

public class Example implements Comparable<Example> {
    private final String id;

    public getId() {
        return id;
    }

    private Integer getIdNo() {
        try {
            return Integer.parseInt(getId().replaceAll("[\\D]", ""));
        } catch (NumberFormatException e) {
            return null;
        }
    }

    @Override
    public int compareTo(Example o) {
        if ((getIdNo() == null && getIdNo() != null) || (getProductFeatureId_sizeNo() < o.getProductFeatureId_sizeNo())) {
            return -1;
        } else if (o.getIdNo() == null || getIdNo() > o.getIdNo()) {
            return 1;
        }

        return 0;
    }
}

Answer 1

这是更好的选择 - AlphanumComparator.java

复制代码以供参考 -

public class AlphanumComparator implements Comparator
{
    private final boolean isDigit(char ch)
    {
        return ch >= 48 && ch <= 57;
    }

    /** Length of string is passed in for improved efficiency (only need to calculate it once) **/
    private final String getChunk(String s, int slength, int marker)
    {
        StringBuilder chunk = new StringBuilder();
        char c = s.charAt(marker);
        chunk.append(c);
        marker++;
        if (isDigit(c))
        {
            while (marker < slength)
            {
                c = s.charAt(marker);
                if (!isDigit(c))
                    break;
                chunk.append(c);
                marker++;
            }
        } else
        {
            while (marker < slength)
            {
                c = s.charAt(marker);
                if (isDigit(c))
                    break;
                chunk.append(c);
                marker++;
            }
        }
        return chunk.toString();
    }

    public int compare(Object o1, Object o2)
    {
        if (!(o1 instanceof String) || !(o2 instanceof String))
        {
            return 0;
        }
        String s1 = (String)o1;
        String s2 = (String)o2;

        int thisMarker = 0;
        int thatMarker = 0;
        int s1Length = s1.length();
        int s2Length = s2.length();

        while (thisMarker < s1Length && thatMarker < s2Length)
        {
            String thisChunk = getChunk(s1, s1Length, thisMarker);
            thisMarker += thisChunk.length();

            String thatChunk = getChunk(s2, s2Length, thatMarker);
            thatMarker += thatChunk.length();

            // If both chunks contain numeric characters, sort them numerically
            int result = 0;
            if (isDigit(thisChunk.charAt(0)) && isDigit(thatChunk.charAt(0)))
            {
                // Simple chunk comparison by length.
                int thisChunkLength = thisChunk.length();
                result = thisChunkLength - thatChunk.length();
                // If equal, the first different number counts
                if (result == 0)
                {
                    for (int i = 0; i < thisChunkLength; i++)
                    {
                        result = thisChunk.charAt(i) - thatChunk.charAt(i);
                        if (result != 0)
                        {
                            return result;
                        }
                    }
                }
            } else
            {
                result = thisChunk.compareTo(thatChunk);
            }

            if (result != 0)
                return result;
        }

        return s1Length - s2Length;
    }
}

注意：如果你使用 java 1.5+，你应该生成这个类

Answer 2

private Integer getIdNo() {
        try {
            return Integer.parseInt(getId().replaceAll("[\\D]", ""));
        } catch (NumberFormatException e) {
            return 0;
        }
    }

@Override
public int compareTo(Example o) {
    if(o == null) {
        return 1;
    }
    return getIdNo().compareTo(o.getIdNo());
}

Answer 3

要在@Jeshuruns 答案的基础上构建，您可以像这样跳过所有数字解析：

  @Override
  public int compareTo(Example o) {
      if (o == null) {
          return 1;
      }
      if (getId().length() != o.getId().length() {
          return (getId().length() - o.getId().length());
      } else {
          return getId().compareTo(o.getId());
      }
  }

Answer 4

你可以写这样的东西。

public class Sorter implements Comparator<String>
{
   public int compareTo(String str1,String str2)
   {
       if( str1 == null )
       {
           return -1;
       }
       if( str2 == null )
       {
           return 1;
       }
       int val1 = Integer.parseInt(this.split("_")[1]);
       int val2 = Integer.parseInt(str.split("_")[1]);

       if(val1<val2)
       {
           return -1;
       }
       else if(val1 > val2)
       {
           return 1;
       }
       else
       {
           return 0;
       }
   }
}