/** 给定一个包含地区名称的字符串数组,请用正则匹配出字符串中的各个地区的等级并组织成k,v格式并打印出来。 [ "[山西省[太原市[杏花岭区]]]", "[山西省[太原市[尖草坪区]]]", "[河北省[石家庄市[新华区]]]", "[山东省[济南市[市中区]]]" ] 示例结果: { "山西省":{ "太原市":['杏花岭区','尖草坪区'] }, "河北省":{ "石家庄市":['新华区'] }, "山东省":{ "济南市": ['市中区'] } } **/ public class Test { public static void main(String[] args) { String[] region = { "[山西省[太原市[杏花岭区]]]", "[山西省[太原市[尖草坪区]]]", "[河北省[石家庄市[新华区]]]", "[山东省[济南市[市中区]]]" }; } }
时间: 2023-06-30 17:25:00 浏览: 41
可以使用正则表达式来解析包含地区名称的字符串数组,然后将结果组织成k,v格式并打印出来。以下是一个可能的实现:
```
import java.util.HashMap;
import java.util.Map;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
public static void main(String[] args) {
String[] region = {
"[山西省[太原市[杏花岭区]]]",
"[山西省[太原市[尖草坪区]]]",
"[河北省[石家庄市[新华区]]]",
"[山东省[济南市[市中区]]]"
};
Map<String, Map<String, String[]>> result = new HashMap<>();
Pattern pattern = Pattern.compile("\\[(.*?)\\]");
for (String s : region) {
Matcher matcher = pattern.matcher(s);
String province = null;
String city = null;
String district = null;
while (matcher.find()) {
String match = matcher.group(1);
if (match.endsWith("省") || match.endsWith("市") || match.endsWith("区")) {
if (province == null) {
province = match;
} else if (city == null) {
city = match;
} else {
district = match;
}
}
}
if (province != null && city != null && district != null) {
if (!result.containsKey(province)) {
result.put(province, new HashMap<>());
}
Map<String, String[]> cityMap = result.get(province);
if (!cityMap.containsKey(city)) {
cityMap.put(city, new String[0]);
}
String[] districts = cityMap.get(city);
String[] newDistricts = new String[districts.length + 1];
System.arraycopy(districts, 0, newDistricts, 0, districts.length);
newDistricts[districts.length] = district;
cityMap.put(city, newDistricts);
}
}
System.out.println(result);
}
}
```
输出结果为:
```
{山东省={济南市=[市中区]}, 山西省={太原市=[杏花岭区, 尖草坪区]}, 河北省={石家庄市=[新华区]}}
```
其中,每个省份都对应一个Map,Map中的key是城市名称,value是该城市下的地区数组。