java - 从列表中删除多个元素范围

Question

My list of objects can have such elements eg1:

[vale11, value12, value13, null, null, null, value21, value22, value23, value31, value32, value33]

eg2:

[vale11, value12, value13, null, null, null, null, null, null, value31, value32, value33]

eg3:

[vale11, value12, value13, null, null, null, value21, value22, value23, null, null, null]

eg4:

[vale11, null, value13, null, null, null, value21, value22, value23, value31, value32, null]

I want to remove the null values but not all (note eg4) and only those in a range started from certain index. So in eg1 would be something like:

list.sublist(3, 6).clear();

eg2:

list.sublist(3, 6).clear();
list.sublist(6, 9).clear();//it's not going to work

I know the starting indexes and a number of next elements (always the same)

Sometimes it would be 1 range, sometimes 3 , 5 ... How to clear the original list with the use of a loop or streams ?

score 2 · Accepted Answer

您可以稍微改变一下您的技术：按降序排列初始索引上的范围。这里的基本原理是，只要您从较高的索引转到较低的索引，并且范围不“交叉”，索引就会保持一致。

因此，清除sublist(6, 9)之后sublist(3, 6)将毫无问题地工作：

//  0        1       2       3     4     5     6     7     8      9       10       11
[vale11, value12, value13, null, null, null, null, null, null, value31, value32, value33]
//                                           ^^^^^^^^^^^^^^^^ [6, 9)
//                         ^^^^^^^^^^^^^^^^ [3, 6)

score 0 · Accepted Answer

使用可以通过循环来做到这一点：

下面的方法将删除其元素startIndex <= index <= endIndex

<T> List<T> chopList(List<T> originalList, int startIndex, int endIndex) {
    // Your code to check original list, validate startIndex, endIndex must inrange...
    return IntStream.range(0, originalList.size() - 1)
                    .filter(i -> i < startIndex || i > endIndex)
                    .mapToObj(originalList::get)
                    .collect(toList());
}

score 0 · Accepted Answer

如果你想null从 a 中删除元素List，你可以这样做很简单

list.removeIf(Objects::isNull);

当然，您可以将操作限制在一定范围内，例如

list.subList(start, end).removeIf(Objects::isNull);

如果实际任务是从列表中删除预定义范围（并且这些位置可能存在null元素的信息实际上是无关紧要的），则可以使用此答案（按降序处理范围），如果范围不重叠.

如果范围可能重叠，您可以使用

BitSet toRemove = new BitSet(list.size());
toRemove.set(firstRangeStart, firstRangeEnd);
toRemove.set(secondRangeStart, secondRangeEnd);
// etc
for(int e = toRemove.length(), s; e > 0; e = toRemove.previousSetBit(s)+1)
    list.subList((s=toRemove.previousClearBit(e-1))+1, e).clear();

这将在按降序处理结果范围之前融合相邻和重叠的范围。

java - 从列表中删除多个元素范围

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