Java正则表达式匹配不到结果的解决

String str = "\uFEFF<?xml version=\"1.0\" encoding=\"utf-8\"?><Response xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\"><Header ShouldRecordPerformanceTime=\"false\" Timestamp=\"2018-6-25 21:24:03\" RequestID=\"2c4d0b24-fd48-4a92-a2d8-c66793df2059\" ResultCode=\"Success\" AssemblyVersion=\"2.9.5.0\" RequestBodySize=\"0\" SerializeMode=\"Xml\" RouteStep=\"1\" Environment=\"pro\" /><SSPATResponse><Result>0</Result><FareDetail /><Price>0</Price><ErrCode>102</ErrCode><DetailInfo>Send:APPLOCK\n" +
    "Rev:\n" +
    "可用资源锁定成功, 60 秒内没有输入指令资源将被Buk收回\n" +
    "Send:IG\n" +
    "Rev:\n" +
    "NO PNR\n" +
    "Send:\n" +
    "SS:AA186/N/27JUN18/PEKORD/NN1;\n" +
    "Rev:\n" +
    "AA 186 N 27JUN PEKORD NN1 WL OPEN \n" +
    "UNABLE TO SELL.PLEASE CHECK THE AVAILABILITY WITH \"AV\" AGAIN\n" +
    "Send:IG\n" +
    "Rev:</DetailInfo><PatOfficeno>SHA717</PatOfficeno></SSPATResponse><ResponseStatus><Timestamp xmlns=\"http://soa.ctrip.com/common/types/v1\">2018-06-25T21:24:03.4535624+08:00</Timestamp><Ack xmlns=\"http://soa.ctrip.com/common/types/v1\">Success</Ack></ResponseStatus></Response>";
 
String regex = "<DetailInfo>((.|\\n")*?)</DetailInfo>";

<DetailInfo>((.|\\n|\\r\\n")*?)</DetailInfo>

String.matches(regex);

Pattern.compile(regex).matcher(str).find()

public boolean matches(String regex) {
    return Pattern.matches(regex, this);
}
Pattern.matches(regex, this)

public static boolean matches(String regex, CharSequence input) {
  Pattern p = Pattern.compile(regex);
  Matcher m = p.matcher(input);
  return m.matches();
}

public static Pattern compile(String regex) {
    return new Pattern(regex, 0);
}
Pattern.matcher

public Matcher matcher(CharSequence input) {
    if (!compiled) {
      synchronized(this) {
        if (!compiled)
          compile();
      }
    }
    Matcher m = new Matcher(this, input);
    return m;
}

public boolean find() {
    int nextSearchIndex = last;
    if (nextSearchIndex == first)
      nextSearchIndex++;
    // If next search starts before region, start it at region
    if (nextSearchIndex < from)
      nextSearchIndex = from;
    // If next search starts beyond region then it fails
    if (nextSearchIndex > to) {
      for (int i = 0; i < groups.length; i++)
        groups[i] = -1;
      return false;
    }
    return search(nextSearchIndex);
}