1、需要：对“1.1”，“1.2”，“1.3”，“1.1.1”，“2”，“4”，“5”，“4.1.1”，“11” 进行排序
想要的效果：“1.1”，“1.1.1”，“1.2”，“1.3”，“2”，“4”，“4.1.1”，“5”，“11”
如果用JAVA原生的compareTo 比较
public static void main(String[] args) throws Exception {
String[] arr = new String[]{"1.1","1.2","1.3","1.1.1","2","4","5","4.1.1","11"};
List<String> list = Arrays.asList(arr);
Collections.sort(list, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
return o1.compareTo(o2);
}
});
System.out.println(list); //[1.1, 1.1.1, 1.2, 1.3, 11, 2, 4, 4.1.1, 5]
}
结果不符合预期。11在2的前面显然不对
2、自制一个算法：也很简单。
1、去除wbsCode里面的小数点，然后再后面补0。转正long类型，进行比较大小。
根据实际实际需求，这里假设，小数点的个数，不会超出9位。
2、在后面补多少个0，要根据wbsCode实际有的小数来判断。补0数 = 9 - wbsCode小数点个数。
public static void main(String[] args) throws Exception {
String[] arr = new String[]{"1.1","1.2","1.3","1.1.1","2","4","5","4.1.1","11"};
List<String> list = Arrays.asList(arr);
Collections.sort(list, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
String wbsCode1 = o1.replaceAll("\\.","");
String wbsCode2 = o2.replaceAll("\\.","");
int pointNum1 = 9-o1.split("\\.").length;
int pointNum2 = 9-o2.split("\\.").length;
String zero1 = "";
String zero2 = "";
for(int index=0;index<pointNum1;index++){
zero1 += "0";
}
for(int index=0;index<pointNum2;index++){
zero2 += "0";
}
wbsCode1 = wbsCode1+ zero1;
wbsCode2 = wbsCode2 +zero2;
// 转换为long类型
long wbsCodeLong1 = Long.parseLong(wbsCode1);
long wbsCodeLong2 = Long.parseLong(wbsCode2);
if(wbsCodeLong1 > wbsCodeLong2){
return 1;
}else{
return -1;
}
}
});
System.out.println(list); //[1.1, 1.1.1, 1.2, 1.3, 2, 4, 4.1.1, 5, 11]
}