在Java中，排序必要复写的是 equals 办法 和 Comparable 接口 的public int compareToT o;。下面是小编为大师带来的Java字符串排序中文和数字的办法，欢送阅读。
办法步调：
1. 使用正则表达式来断定数字，多个连气儿的数字作为一组,
2. 一次检索出数字组合，
3. 检出下一组数字,假如有,则进入步调4，不然进入步调6.
4. 假如两组数字呈现的地位相等，而且后面局部的字符串相等，则进入第5步。不然break，跳到第6步.
5. 假如后面局部的字符串完全分歧。则比拟两个数字的大小，假如大小分歧，则进入下一组，即步调3.假如大小纷歧致，则可以比对出来大小，比拟结束
6. 调用String的compareTo办法,病返回流程结束。
完整的'代码如下：
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
//包装器类
public class OrderWrapper implements Comparable
String name = null;
public OrderWrapperString name
this.name = name;
public String getName
return name;
public void setNameString name
this.name = name;
@Override
public String toString
return String.valueOfname;
@Override
public boolean equalsObject obj
ifobj == this
return true;
ifobj instanceof OrderWrapper
OrderWrapper other = OrderWrapperobj;
ifnull == this.name
return false;
else
return this.name.equalsother.name;
return false;
// 比拟办法,相当于减法。 return this - wrapper
public int compareToOrderWrapper wrapper
ifnull == wrapper
return 1;
// 间接相等
ifthis == wrapper || this.equalswrapper
return 0;
String name1 = this.name;
String name2 = wrapper.name;
// 非凡情形，name有一个为空的环境.
ifnull == name1
// 都为空，认为相对于
ifnull == name2
return 0;
else
return -1;
else ifnull == name2
return 1;
// 中间 1-多个数字
Pattern pattern = Pattern.compile"D*d+D*";
Matcher matcher1 = pattern.matchername1;
Matcher matcher2 = pattern.matchername2;
//System.out.printlnpattern.pattern;
//
int index1_step = 0;
int index2_step = 0;
whilematcher1.find
String s1 = matcher1.group1;
String s2 = null;
ifmatcher2.find
s2 = matcher2.group1;
int index1 = name1.indexOfs1, index1_step;
int index2 = name2.indexOfs2, index2_step;
//
index1_step = index1;
index2_step = index2;
// 索引相等的环境下
ifindex1 == index2
System.out.println"name1="+name1.length+"nname2="+name2.length;
System.out.println"index1="+index1+",index2="+index2;
String pre1 = name1.substring0, index1;
String pre2 = name2.substring0, index2;
ifpre1.equalspre2
//
long num1 = Long.parseLongs1;
long num2 = Long.parseLongs2;
//
ifnum1 == num2
// 比拟下一组
continue;
else
return intnum1 - num2;
else
break;
else
break;
// 最后的情形.
return this.name.compareTowrapper.name;
public static void testNew
List chinesesOrderList = new ArrayList;
chinesesOrderList.addnew OrderWrapper"我们80后相亲奇遇记-1.mp3";
chinesesOrderList.addnew OrderWrapper"他80后相亲奇遇记-10.mp3";
chinesesOrderList.addnew OrderWrapper"我80后相亲奇遇记-11.mp3";
chinesesOrderList.addnew OrderWrapper"啊80后相亲奇遇记-12.mp3";
chinesesOrderList.addnew OrderWrapper"我80后相亲奇遇记-13.mp3";
chinesesOrderList.addnew OrderWrapper"我80后相亲奇遇记-25.mp3";
chinesesOrderList.addnew OrderWrapper"我80后相亲奇遇记-26.mp3";
chinesesOrderList.addnew OrderWrapper"我80后相亲奇遇记-2.mp3";
chinesesOrderList.addnew OrderWrapper"我80后相亲奇遇记-3.mp3";
chinesesOrderList.addnew OrderWrapper"我80后相亲奇遇记-4.mp3";
chinesesOrderList.addnew OrderWrapper"a80后相亲奇遇记-4.mp3";
//Collator collatorChinese = Collator.getInstancejava.util.Locale.CHINA;
//collatorChinese = Collator.getInstancejava.util.Locale.CHINESE;
// Collections.sortchinesesOrderList, collatorChinese;
Collections.sortchinesesOrderList;
System.out.println"中文+数字排序： = ";
for int i = 0; i < chinesesOrderList.size; i++
OrderWrapper chinese = chinesesOrderList.geti;
System.out.println"" + chinese;
public static void mainString[] args
testNew;