Question

我遇到了一些困难。我想调整下面的这段代码来对Person类型的单链表进行排序。例如，我有：

class Person{
    private String fn = "NN";
    private String ln = "NN";
    private Date dob = new Date(1, 1, 1970);
}

我想按名字，姓氏和出生日期对人员的链接进行排序。与此同时，我获得了一段代码以适应它，但我似乎无法找到解决方法。任何帮助将受到高度赞赏。以下是适应的代码：

public void sort() {

    TextIO.putln("sorting...");
    if (head == null)
        return;

    Node max, t, out = null, h = new Node();
    h.next = head;

    while (h.next != null) {
        // display(); 
        TextIO.putln("max=" + (max = findmax(h)).next);
        t = max.next;
        max.next = t.next;
        t.next = out;
        out = t;
    }
    head = out;

}

private Node findmax(Node h) {
    char max = '\0';
    Node maxNode = h;
    for (Node tmp = h; tmp.next != null; tmp = tmp.next) {
        if (tmp.next.data.c >= max) {
            max = tmp.next.data.c;
            maxNode = tmp;
        }
    }
    return maxNode;
}

如果没有，详细的建议将非常有用，谢谢。请注意，我不能使用collection.sort或任何其他就绪函数，它必须实现。

由于

Answer 1

首先，我希望你看看this link。它将告诉您如何实现Node类以及如何实现列表。

假设您从链接中读取内容，这里有一些评论;

public void sort() { 
     //this method does ascending sort of the list

     TextIO.putln("sorting..."); 
     if (head == null) 
         return; 

     Node max, t, out = null, h = new Node(); 
     h.next=head; //make new list node let him point to beginning of the list(head)

     while (h.next != null) { //until we get to the end of the list
         // display(); 

         TextIO.putln("max="+(max = findmax(h)).next);
         //after findmax is finished we know reference
         //to the node that contains max value
         //and we need to bring that node to the beginning of the list
         //that requires some reference rewiring
         //first set t to point to next node from max
         t = max.next;
         //than max node will point to the next node from t
         //this way max node becomes detached from list
         max.next = t.next;
         // now max node will point to some node that will represent new list head
         // not new list just new list head
         t.next = out; 
         out = t; 
     } 
     //after we complete sorting just point list head to the sorted list
     head = out; 

 } 

 //find node that contains max value starting from some node   
 private Node findmax(Node h) { 
     //declare that max is null char
     char max = '\0'; 
     // set maxNode to be current node
     Node maxNode = h; 
     //go through all nodes starting from current which is h
     for (Node tmp = h; tmp.next != null; tmp = tmp.next) { 
         //if the data of some node in the list is bigger then our max value
         if (tmp.next.data.c >= max) {
             //then we are going to set that new value to be max 
             max = tmp.next.data.c;
             // and set maxNode to be the node that has max value
             maxNode = tmp; 
         } 
     } 
     return maxNode; 
 }

我的建议是查看提供的链接并开始使用好的旧笔和纸。这是理解列表指针和节点发生了什么的最好方法。但如果你不想要笔和纸，你可以去this link，它将显示基本列表操作的动画。

按名称和出生日期对链接的链接列表进行排序

1 个答案: