连接具有不同最后定界符的字符串

如果它们已经在列表中，则不需要流。只需加入除最后一个元素以外的所有元素的子列表，并连接另一个定界符和最后一个元素：

int last = list.size() - 1;String joined = String.join(" and ",         String.join(", ", list.subList(0, last)),         list.get(last));

这是使用上述功能的版本

Collectors.collectingAndThen:

stream.collect(Collectors.collectingAndThen(Collectors.toList(),    joiningLastDelimiter(", ", " and ")));public static Function<List<String>, String> joiningLastDelimiter(        String delimiter, String lastDelimiter) {    return list -> {     int last = list.size() - 1;     if (last < 1) return String.join(delimiter, list);     return String.join(lastDelimiter,         String.join(delimiter, list.subList(0, last)),         list.get(last)); };}

此版本还可以处理流为空或只有一个值的情况。感谢Holger和Andreas的建议，这些建议极大地改善了此解决方案。

我曾在评论中建议，可以使用

", "

", and"

"a, andb"

stream.collect(Collectors.collectingAndThen(Collectors.toList(),    joiningOxfordComma()));public static Function<List<String>, String> joiningOxfordComma() {    return list -> {     int last = list.size() - 1;     if (last < 1) return String.join("", list);     if (last == 1) return String.join(" and ", list);     return String.join(", and ",         String.join(", ", list.subList(0, last)),         list.get(last)); };}