方法

The method Collections.indexOfSubList will give you the desired information.

返回指定源列表中指定目标列表第一次出现的起始位置，如果没有出现，则返回-1. 更正式地，返回最低索引i，这样source.subList(i，i + target.size()).equals(target)或-1(如果没有这样的索引). (如果target.size()> source.size()，则返回-1.)

Returns the starting position of the first occurrence of the specified target list within the specified source list, or -1 if there is no such occurrence. More formally, returns the lowest index i such that source.subList(i, i+target.size()).equals(target), or -1 if there is no such index. (Returns -1 if target.size() > source.size().)

int index=Collections.indexOfSubList(parentDataList, child1);
…

索引间隔将从index(含)到index+child1.size()(不含).当然，除非返回的索引是-1.在后一种情况下，找不到子列表.

The index interval will be from index, inclusive, to index+child1.size(), exclusive. Unless the returned index is -1, of course. In the latter case the sublist was not found.