此字符串的输出应按以下顺序排序。 该字符串的输出应按以下顺序排序。
public static void main(String[] args) {
String input="";
List<String> items = Arrays.asList(input.split("\\s*,\\s*"));
System.out.println("items: " + items);
Collections.sort(items, new Comparator<String>() {
public int compare(String o1, String o2) {
String o1StringPart = o1.replaceAll("\\d", "");
String o2StringPart = o2.replaceAll("\\d", "");
if (o1StringPart.equalsIgnoreCase(o2StringPart)) {
return extractInt(o1) - extractInt(o2);
}
return o1.compareTo(o2);
}
int extractInt(String s) {
String num = s.replaceAll("\\D", "");
// return 0 if no digits found
return num.isEmpty() ? 0 : Integer.parseInt(num);
}
});
for (String s : items) {
System.out.println(s);
}} }
答案 0 :(得分:0)
我将提供一些提示
为了比较项目1,20,200-2C,3,32C,32D,4(这是默认的排序顺序,作为字符串),您需要检查字符串中是否包含任何数字。可以使用正则表达式来完成。您可以在网上找到很多带有正则表达式的示例,这些正则表达式可以带回字符串中的数字。
修改比较器,并检查要比较的两个字符串中的任何一个是否包含任何数字,并返回适当的结果。
extractInt方法可以类似于@Pankaj Saini的建议
Integer extractInt(String s) {
Pattern pattern = Pattern.compile("^\\d+");
Matcher matcher = pattern.matcher(s);
if (matcher.find()) {
String num = matcher.group(0);
return Integer.parseInt(num);
}
return null;
}
compare方法可以像这样
public int compare(String o1, String o2) {
if(extractInt(o1)!=null && extractInt(o2)!=null){
if(extractInt(o1).equals(extractInt(o2)))
{
return o1.substring(extractInt(o1).toString().length())
.compareTo(o2.substring(extractInt(o2).toString().length()));
}
return extractInt(o1).compareTo(extractInt(o2));
}
else if(extractInt(o1)!=null)
{
return -1;
}
else if(extractInt(o2)!=null)
{
return 1;
}
else{
return o1.compareTo(o2);
}
}
答案 1 :(得分:0)
使用以下代码段代码。首先,需要比较字符串的整数部分,如果整数部分相等,则比较字符串部分。
import java.io.*;
import java.util.*;
import java.util.regex.*;
/*
* To execute Java, please define "static void main" on a class
* named Solution.
*
* If you need more classes, simply define them inline.
*/
class Solution {
public static void main(String[] args) {
String input = "605,2A,401-2A,32C,21F,201A,605A,401-1A,200-2E,583-58D,583/58E,583-57D,542,2B,1,542/2E,605B,32D,3,603,4,6,5,60,201C,542/2D,40,20,50,200-2C,21C,800A,200A,571-573B,51/2,470/1,51/1,571-573C,454-1,444-446";
List < String > items = Arrays.asList(input.split("\\s*,\\s*"));
System.out.println("items: " + items);
Pattern pattern = Pattern.compile("^\\d+");
Collections.sort(items, new Comparator < String > () {
public int compare(String o1, String o2) {
int intDiff = extractInt(o1) - extractInt(o2);
if (intDiff == 0) {
return o1.compareTo(o2);
}
return intDiff;
}
int extractInt(String s) {
Matcher matcher = pattern.matcher(s);
if (matcher.find()) {
String num = matcher.group(0);
return Integer.parseInt(num);
}
return 0;
}
});
for (String s: items) {
System.out.println(s);
}
}
}
答案 2 :(得分:0)
我假设所有数字都是整数,并且所有字母都是A到Z,您可以先将包含/
或-
的字符串转换为浮点,然后将所有字母替换为空字符串。最后,将它们的值比较为Double
。
例如,51/1
将是51.1
,而571-573B
将是571.573
。
代码段
public int compare(String o1, String o2) {
String n1 = o1.replace("-", ".").replace("/", ".").replaceAll("[A-Z]", "");
String n2 = o2.replace("-", ".").replace("/", ".").replaceAll("[A-Z]", "");
// This equals above statements
//String n1 = o1.replaceAll("[-/]", ".").replaceAll("[A-Z]", "");
//String n2 = o2.replaceAll("[-/]", ".").replaceAll("[A-Z]", "");
int result = Double.compare(Double.valueOf(n1), Double.valueOf(n2));
return (result == 0) ? o1.compareTo(o2) : result;
}
这不是最优雅的方法,但是我认为它应该起作用!
答案 3 :(得分:0)
此答案处理20-10A和20-2A之类的数字之间的比较
public static void main(String[] args) {
String s = "605,2A,401-2A,32C,21F,201A,605A,401-1A,200-2E,583-58D,583/58E,583-57D,542,2B,1,542/2E," +
"605B,32D,3,603,4,6,5,60,201C,542/2D,40,20,50,200-2C,21C,800A,200A,571-573B,51/2,470/1,51/1," +
"571-573C,454-1,444-446";
String[] strings = s.split(",");
Arrays.sort(strings, App::compare);
System.out.println(Arrays.deepToString(strings));
}
public static int compare(String o1, String o2) {
if (startsWithDelim(o1)) return compare(o1.substring(1), o2);
if (startsWithDelim(o2)) return compare(o1, o2.substring(1));
List<String> n1 = extractInt(o1);
List<String> n2 = extractInt(o2);
if (n1 != null && n2 != null) {
Integer n1int = Integer.parseInt(n1.get(0));
Integer n2int = Integer.parseInt(n2.get(0));
String n1Remaining = n1.get(1);
String n2Remaining = n2.get(1);
int intCompare = n1int.compareTo(n2int);
return intCompare == 0 ? compare(n1Remaining, n2Remaining) : intCompare;
}
if (n1 == null && n2 == null)
return o1.compareTo(o2);
else if (n1 == null) return -1;
else return 1;
}
static List<String> extractInt(String s) {
Pattern pattern = Pattern.compile("^\\d+");
Matcher matcher = pattern.matcher(s);
if (matcher.find()) {
String num = matcher.group(0);
return List.of(num, s.substring(matcher.end(0)));
}
return null;
}
static boolean startsWithDelim(String s) {
return (s.startsWith("/") || s.startsWith("-"));
}