使链接列表唯一 [英] Making a linked list unique

问题描述

所以我有列表:

 8,9,1,2,3,5,3,1,8,1,4,5,2,7,9,9,5,6

，并希望将其复制为唯一的(仅包含相同数字之一)，因此它将是

and want to reproduce it as unique (only containing one of the same number) so this one would be

8,9,1,2,3,5,4,7,6

发生的唯一问题是remove()方法删除了节点值的第一个实例，并且假定要保留一个数字的第一个实例，并假定要从列表中删除以下实例.没有集合，哈希图或其他任何东西.

the only issue that is occurring is that the remove() method deletes the first instance of a nodes value and the first occurances of a number are suppose to be kept and following instances are suppose to be deleted from the list. Without collections, hashmaps, or anything.

    while(current.getPrevious() != null) {
        while(next.getPrevious() != null) {
            if(next.equals(current)){
                next = next.getNext();
                continue;
            }
            else if(next.getValue().compareTo(current.getValue()) == 0){
                remove(next.getValue()); // The remove method deletes the first instance of number in the linked list
                next = next.getNext();

            }
                next = next.getPrevious();

        }
        current = current.getNext();
        next = current.getNext();
    }

推荐答案

将其全部扔到 LinkedHashSet 中，这将保留顺序并仅保留第一个副本.

Throw it all in a LinkedHashSet which will keep the order preserved and only keep the first copy.

Set<Integer> noDups = new LinkedHashSet<Integer>(myList);

如果您将列表与 System.out.println(noDups); 一起使用，则为以下结果:

Here is the result if you use your list with System.out.println(noDups); :

[8, 9, 1, 2, 3, 5, 4, 7, 6]

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