如何从 Arraylist 中获取唯一项，保持自然顺序 [英] how to get unique items from the Arraylist, preserving the natural order

问题描述

例如 ArrayList :

For example ArrayList :

 Integer[] intArray = new Integer[] { 0, 1 , 0 , 2 , 3 , 3 , 5 , 6 , -4 ,6 };
 ArrayList<Integer> list = new ArrayList<Integer>(Arrays.asList(intArray));

需要获取非重复值，保持顺序

need to get non-repeating values, preserving order

1 , 2 , 5 , -4

0、3 和 6 被删除，因为它们出现了不止一次.

0, 3 and 6 are removed because they occur more than once.

推荐答案

将元素放入LinkedHashSet;如果你找到一个已经存在的元素，把它放到另一个集合中；删除最后所有已经存在"的元素:

Put the elements into a LinkedHashSet; if you find an element that's already present, put it into another set; remove all the "already present" elements at the end:

LinkedHashSet<Integer> set = new LinkedHashSet<>();
HashSet<Integer> alreadyPresent = new HashSet<>();
for (Integer i : intArray) {
  if (!set.add(i)) {
    alreadyPresent.add(i);
  }
}
set.removeAll(alreadyPresent);

ArrayList<Integer> list = new ArrayList<>(set);

Ideone 演示

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