“我有最短的”挑战，以下是我的;)

迭代:

public static String join(String s, Object... a) {
    StringBuilder o = new StringBuilder();
    for (Iterator<Object> i = Arrays.asList(a).iterator(); i.hasNext();)
        o.append(i.next()).append(i.hasNext() ? s : "");
    return o.toString();
}

递归:

public static String join(String s, Object... a) {
    return a.length == 0 ? "" : a[0] + (a.length == 1 ? "" : s + join(s, Arrays.copyOfRange(a, 1, a.length)));
}

要从 “str1”、“str2”， “” 获取 “str1， str2”：

Stream.of("str1", "str2", "").filter(str -> !str.isEmpty()).collect(Collectors.joining(", ")); 

此外，您还可以添加额外的空检查

在Java 8中你可以使用

1)流API:

String[] a = new String[] {"a", "b", "c"};
String result = Arrays.stream(a).collect(Collectors.joining(", "));

2) new String。加入方式:https://stackoverflow.com/a/21756398/466677

3) java.util.StringJoiner类:http://docs.oracle.com/javase/8/docs/api/java/util/StringJoiner.html

用一个小mod代替substring():

//join(String array,delimiter)
public static String join(String r[],String d)
{
        if (r.length == 0) return "";
        StringBuilder sb = new StringBuilder();
        int i;
        for(i=0;i<r.length-1;i++){
            sb.append(r[i]);
            sb.append(d);
        }
        sb.append(r[i]);
        return sb.toString();
}

你可以很容易地在大约十行代码中编写这样一个函数:

String combine(String[] s, String glue)
{
  int k = s.length;
  if ( k == 0 )
  {
    return null;
  }
  StringBuilder out = new StringBuilder();
  out.append( s[0] );
  for ( int x=1; x < k; ++x )
  {
    out.append(glue).append(s[x]);
  }
  return out.toString();
}

如果你正在使用Spring框架，那么你有StringUtils类:

import static org.springframework.util.StringUtils.arrayToDelimitedString;

arrayToDelimitedString(new String[] {"A", "B", "C"}, "\n");