链接列表连接

Question

为分配创建链接列表和一个需求是一个名为concat的方法,它接受一个list参数并将其附加到当前列表的末尾.这个方法没有必要使用递归,但我们的程序应该大量使用递归.我只是想知道是否有可能为此提出递归算法.我们的列表类只有一个头节点,没有别的,它们没有双重链接.

我当前的尝试只能递归地附加第一个值.我知道它做错了什么,但我无法想出解决方案.第一种方法是在列表中实际调用的方法,其中列表被传递到"连接".然后我尝试找到列表的尾部并将它们传递给递归方法.这种"包装器"方法是我们递归方法的强制要求.这是我的尝试,但显然失败了,因为一旦所有调用从堆栈中弹出然后重新进入concat的递归调用,我就无法将节点引用"pt"推进到列表中的下一个节点.如果这是可能的递归,可以请你给我一个如何推进这一价值减记第一个列表,重新进入递归调用或可能对这个问题只是一个更好的一般方法的想法？谢谢你的时间.

public void concat(MyString list1) {
    CharacterNode tail = null, pt = list1.head;
    // Find the tail of the list
    if (pt == null) {
    } else if (pt.getNext() == null) {
        tail = pt;
    } else {
        while (pt.getNext() != null) {
            pt = pt.getNext();
        }
        tail = pt;
    }
    list1.head = concat(list1.head, tail, list1.head);
}
private static CharacterNode concat(CharacterNode lhead, CharacterNode tail, CharacterNode pt) {
    // Pass in smaller list every time
    // Head traverses down list till the end
    // Add new node with (pt's letter, null link)
    if (lhead == null) {
    // If head is null, we need to add the node
        lhead = new CharacterNode(pt.getCharacter(),null);
    } else if (tail.getNext() == lhead) {
    // If the head is past tail, stop   
    } else {
        // Call concat on a smaller list
        lhead.setNext(concat(lhead.getNext(),tail,pt));
    }
    return lhead;
}

这是CharacterNode:

class CharacterNode {
    private char letter;
    private CharacterNode next;

    public CharacterNode(char ch, CharacterNode link) {
        letter = ch;
        next = link;
    }

    public void setCharacter(char ch) {
        this.letter = ch;
    }

    public char getCharacter() {
        return letter;
    }

    public void setNext(CharacterNode next) {
        this.next = next;
    }

    public CharacterNode getNext() {
        return next;
    }
}

MyString的:

class MyString {
    // member variable pointing to the head of the linked list
    private CharacterNode head;

    // default constructor
    public MyString() {
    }

    // copy constructor
    public MyString(MyString l) {
    }

    // constructor from a String
    public MyString(String s) {
    }

    // for output purposes -- override Object version
    // no spaces between the characters, no line feeds/returns
    public String toString() {
    }

    // create a new node and add it to the head of the list
    public void addHead(char ch) {
    }

    // create a new node and add it to the tail of the list -- "wrapper"
    public void addTail(char ch) {
    }

    // Recursive method for addTail
    private static CharacterNode addTail(CharacterNode L, char letter) {
    }

    // modify the list so it is reversed
    public void reverse() {
    }

    // remove all occurrences of the character from the list -- "wrapper"
    public void removeChar(char ch) {
    }

    // Recursive removeChar method
    private static CharacterNode removeChar(CharacterNode n, char letter) {
    }

    // "wrapper" for recursive length()
    public int length() {
    }

    // Returns the length of the linked list
    private static int length(CharacterNode L) {
    }

    // concatenate a copy of list1 to the end of the list
    public void concat(MyString list1) {
    }

    // recursive method for concat
    private static CharacterNode concat(CharacterNode lhead, CharacterNode tail, CharacterNode pt) {
    }
}

Answer 1

要连接两个链接列表,必须使第一个列表的最后一个节点指向第二个列表的第一个节点.

Node first_list = ...  // head node
Node second_list = ... // head node
...
Node last_node = first_list.getLastNode()
last_node.setNext(second_list)

现在集中精力实施getLastNode().它可以通过使用递归或迭代非常简单地完成,字面上是2行.