Question

当我跟踪lastC（链接列表中的最后一个节点）时，会出现问题。由于某种原因，在方法之外它似乎一直是空的。我想看看是否有更好的方法来跟踪它以改善我的程序的运行时间。

附加：

public MyStringBuilder append(String s)
{

    if(this.length == 0) {
        firstC = new CNode(s.charAt(0));
        length = 1;
        CNode currNode = firstC;
        // Create remaining nodes, copying from String.  Note
        // how each new node is simply added to the end of the
        // previous one.  Trace this to see what is going on.
        for (int i = 1; i < s.length(); i++)
        {
            CNode newNode = new CNode(s.charAt(i));
            currNode.next = newNode;
            currNode = newNode;
            length++;
        }
        lastC = currNode;
        //System.out.println(lastC.data);
        //lastC.next = null;//new
    } else {
        if(lastC == null) {
            lastC = getCNodeAt(length - 1);
            //System.out.println("no");
        }
        //System.out.println(lastC.data);
        //CNode currNode = lastC;
        //CNode currNode = firstC;//changed
        //for(int j = 0; j < length - 1; j++) {
        //  currNode = currNode.next;
        //}
        CNode currNode = lastC;
        //lastC = currNode;
        for (int i = 0; i < s.length(); i++)
        {

            CNode newNode = new CNode(s.charAt(i));
            currNode.next = newNode;
            currNode = newNode;//
            length++;
        }
        lastC = currNode;


    } 
    //lastC = getCNodeAt(length - 1);
    return this;
}

构造函数：

public MyStringBuilder(String s)
    {
        if (s == null || s.length() == 0) // Special case for empty String
        {                                 // or null reference
            firstC = null;
            lastC = null;
            length = 0;
        }
        else
        {
            // Create front node to get started
            firstC = new CNode(s.charAt(0));
            length = 1;
            CNode currNode = firstC;
            // Create remaining nodes, copying from String.  Note
            // how each new node is simply added to the end of the
            // previous one.  Trace this to see what is going on.
            for (int i = 1; i < s.length(); i++)
            {
                CNode newNode = new CNode(s.charAt(i));
                currNode.next = newNode;
                currNode = newNode;
                length++;
            }
            lastC = currNode;
            //lastC = getCNodeAt(length - 1);
            //System.out.println(lastC.data);
        }
    }

实例变量

private CNode firstC;   // reference to front of list.  This reference is necessary
                            // to keep track of the list
    private CNode lastC;    // reference to last node of list.  This reference is
                            // necessary to improve the efficiency of the append()
                            // method
    private int length;

Answer 1

建议：

使用sentinel节点摆脱不必要的if s
传统上，append()应为void
构造函数调用append()，因为他们正在执行类似的工作

不是Java程序员，但这样的事情应该有效：

public class MyStringBuilder {
    private CNode sentinel = new CNode('s');
    private CNode lastC = sentinel;
    private int length = 0;

    public MyStringBuilder(String s){
        append(s);
    }

    public void append(String s){
        for (int i = 1; i < s.length(); i++){
            CNode newNode = new CNode(s.charAt(i));
            lastC.next = newNode;
            lastC = newNode;
            length++;
        }
    }

    public CNode getHead(){
        return sentinel.next;
    }

    public CNode getTail(){
        if (length) {
            return lastC;
        } else {
            return null;
        }
    }
}

最后一个节点未被跟踪，链接列表Java

1 个答案: