使用Java 8从列表中有选择地删除前N个 [英] Remove first N selectively from list using java 8

问题描述

查看问题注释.有选择地删除N个元素(条件是列表元素匹配删除")

See comments for question. Remove N elements selectively (Condition is that list element matches 'remove')

List<String> mylist = new ArrayList<>();
mylist.add("remove");
mylist.add("all");
mylist.add("remove");
mylist.add("remove");
mylist.add("good");
mylist.add("remove");

//  Remove first X "remove".
//  if X is 2, then result list should be "all, remove, good, remove"
//  Use java 8 features only, possibly single line code.
//  Please don't answer with looping, iterating, if conditions etc.
//  Answer should use JDK 8 new features.

推荐答案

这是怎么回事:

List<String> filter(List<String> mylist, int x){
    AtomicInteger index = new AtomicInteger(0);
    mylist.removeIf(p -> p.equals("remove") && index.getAndIncrement() < x);
    return myList;
}

当x = 0时，它会打印:

With x=0, it prints:

[全部删除，删除，删除，良好，删除]

[remove, all, remove, remove, good, remove]

使用x = 1时，它会打印:

With x=1, it prints:

[全部，删除，删除，良好，删除]

[all, remove, remove, good, remove]

x = 2时，它会打印:

With x=2, it prints:

[全部，删除，良好，删除]

[all, remove, good, remove]

在x = 3的情况下，它会打印:

With x=3, it prints:

[所有，很好，删除]

[all, good, remove]

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