复制单链接列表 [英] duplicate the singly linked list

问题描述

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

[3, 5, 4, 2, 1] where I need the delete the nodes close to the tail which is it should be like [1, 2, 3, 5, 4] any suggestions?

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

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

推荐答案

你不需要删除任何东西（我的意思是不是通过调用 removeValue ）。只需将您遇到的值存储在集合中，并且如果该值已经在集合中，则会重新链接您的列表。如果您没有使用库代码的权利，请使用二叉搜索树实现您的设置，这将非常简单而且非常有效。

You don't need to remove anything at all (I mean not by calling removeValue). Just store the values you encounter in a set and if the value is already in the set, re-link your list in consequence. If you don't have the right to use library code, implement your set with a binary search tree, it will be easy and quite efficient.

这是我将如何做，假设我执行 Set ：

This is how I would do, assuming I have an implementation of Set :

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();
    }
}

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