Question

[3,5,4,2,1]我需要删除靠近尾部的节点，它应该像[1,2,3,5,4]任何建议吗？

public void delete() {
    for(Node<T> current = getHead(); current != null; current = current.getNext()){

                System.out.println(temp.getValue());
                removeValue(temp.getValue());
            }
        }
    }
}

Answer 1

您根本不需要删除任何内容（我的意思是不是通过调用removeValue）。只需将您遇到的值存储在一个集合中，如果该值已在集合中，则重新链接您的列表。如果您没有权利使用库代码，请使用二叉搜索树实现您的集合，这将非常简单且高效。

这就是我要做的，假设我有Set的实现：

public void makeUnique() {
    Set<T>  set      = new MySet<>();
    Node<T> current  = getHead();
    Node<T> previous = null;
    while (current != null) {
        // if current.getValue() is already in the set, we skip this node
        // (the add method of a set returns true iff the element was not 
        // already in the set and if not, adds it to the set)
        if (set.add(current.getValue()) { 
            // previous represents the last node that was actually inserted in the set.
            // All duplicate values encountered since then have been ignored so 
            // previous.setNext(current) re-links the list, removing those duplicates
            if (previous != null) {
                previous.setNext(current); 
                current.setPrevious(previous);
             }
            previous = current;
        }
        current = current.getNext();
    }
}

复制单链表

1 个答案: