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