Java带数字的排序算法字符串 [英] Java Sort algorithm String with Number

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本文介绍了Java带数字的排序算法字符串的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一个字符串列表。 Example_1,EXAMPLE_2,EXAMPLE_3 ... EXAMPLE_99
在这里排序的最佳算法是什么?

I have a List of Strings. EXAMPLE_1, EXAMPLE_2, EXAMPLE_3 ... EXAMPLE_99 What is the best algorithm for sorting here?

使用整理程序是否可能?
这是我目前的程序,但我想可能会有更好的方法:

Is it possible with a Collator? This is my current procedure, but I guess there could be a better way:

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;
    }
}


推荐答案

此是更好的选择- 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;
    }
}

请注意:如果您正在使用Java 1.5 +

这篇关于Java带数字的排序算法字符串的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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