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java - 如何对链表节点数组进行从高到低排序(基于float值)

转载 作者:行者123 更新时间:2023-12-02 13:38:10 24 4
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我有一个链表节点数组,对应于学生注册的类(class),我需要将成绩从最高到最低排序。我花了一整天的时间尝试不同的排序算法,但我似乎无法让它正常工作。诚然,我不太擅长排序算法,但这并不是因为缺乏尝试。我第一次尝试合并排序,但失败得很惨,而且我最近的努力也没有奏效。由于它是一个链表节点数组,因此我对如何实现排序感到困惑,因为大多数算法都针对数组或链表,而不是两者。

我遇到的问题:

1.) 多次插入后显示重复节点

2.) 在节点之间交换等级值,而不是交换节点本身

3.) 根本没有显示任何数据。

如果有人能给我指出正确的方向或帮助解释如何正确排序我的节点,我将不胜感激。我已经阅读了几个小时但没有成功。

当我不对节点进行排序时,节点链接良好并显示良好,因此我知道功能运行正常。我的类 Course() 使用包装类 AGradeListNode() 来保存节点插入数组所需的数据。我的代码中包含我最近对节点进行排序的尝试(这不起作用,导致重复)。请不要对我大喊大叫,我真的在努力学习......

public class Course implements Serializable
{
private String courseName;
private String classTime;
private int courseID;
private Instructor instructor;
private int numStudents; //variable for number of students in the list
private int hold; //used to hold location of node
protected static final int NUL = -1; //Constant class var for end of list symbol.
protected int currentPos; //Current position for iteration
protected int list; //Reference to the first node on the list
protected int free; //Reference to the first node on the free list
protected AGradeListNode[] nodes; // Array of AListNode holds the linked list

// set by find method
protected boolean found; // true if element found, else false
protected int location; // node containing element, if found
protected int previous; // node preceding location

//wrapper class to create nodes for linked list array
protected class AGradeListNode
{
private float grade;
private Student student;
private int next; //a link to the next node in linked list array

//displays nodes as neatly formatted string
@Override
public String toString()
{
return "Student: " + student.getLastName() + ", " + student.getFirstName() + "\n" +
"Grade: " + grade + "\n";
}
}
//constructor initializes array of nodes with 2 lists: used and free.
//contains other data needed for Course object instantiation as well
public Course(int maxNumStudents, String courseName, String classTime, int courseID, Instructor instructor)
{
nodes = new AGradeListNode[maxNumStudents];

for (int i = 0; i < maxNumStudents; i++)
{
nodes[i] = new AGradeListNode();
}

//link together the free nodes
for (int i = 1; i < maxNumStudents; i++)
{
nodes[i - 1].next = i;
}

nodes[maxNumStudents - 1].next = NUL;

list = NUL;
free = 0;
numStudents = 0;
currentPos = NUL;

this.courseName = courseName;
this.classTime = classTime;
this.courseID = courseID;
this.instructor = instructor;
}
// Returns the index of the next available node from the free list
// and updates the free list index
protected int getNode()
{
hold = free;
free = nodes[free].next;
return hold;
}

//Frees the node at array position index by linking it into the
//free list
protected void freeNode(int index)
{
nodes[index].next = free;
free = index;
}
//here is where im having issues. I'm attempting to sort the nodes based on
//grade from high to low. Each grade belongs to a specific student object
public void addStudentGrade(Student stu, float grade)
{
if (isFull())
{
return;
}

int newNode = getNode();
nodes[newNode].student = stu;
nodes[newNode].grade = grade;
nodes[newNode].next = list;
list = newNode;
numStudents++;

//sorting list, not working :/
for (int j = 1; j <= numStudents; j++)
{
float hold = nodes[j].grade;
int i = j-1;
while ((i > -1) && (nodes[i].grade > hold))
{
nodes[i+1] = nodes[i];
i--;
}
nodes[i+1] = nodes[j];
}

for (int i = 0; i < numStudents; i++)
{
System.out.println(nodes[i].toString());
}
}

编辑:这是我根据 Andreas 的宝贵意见更新的 addStudentGrade(Student Stu, float Grade) 方法。不过,我的列表仍然没有正确排序。成绩是根据输入的顺序而不是降序/升序进行排序的(老实说,这对我来说都很好,我只想对列表进行排序:/)。如果有人能发现我做错了什么并帮助我,我将非常感激。我已经修改这个方法三天了,我开始变得非常灰心:

//Method used to add student grades to nodes array. After LL node
//is added, method sorts the used nodes in descending order based
//off the float value grade
public void addStudentGrade(Student stu, float grade)
{
if (isFull())
{
return;
}

int newNode = getNode();
nodes[newNode].student = stu;
nodes[newNode].grade = grade;
nodes[newNode].next = list;
list = newNode;
numStudents++;

//storing used nodes in temp Integer array
Integer[] usedNodes = new Integer[numStudents];

for (int i = list, j = 0; i != NUL; i = nodes[i].next)
{
usedNodes[j++] = i;
}

//sorting the temp array using lambda expression and
//Arrays.sort. I believe this is where my mistake is
Arrays.sort(usedNodes, (i1, i2) ->
{
return Float.compare(nodes[i2].grade, nodes[i1].grade);
});

//rebuilding the linked list after sorting
list = usedNodes[0];
for (int i = 1; i < usedNodes.length; i++)
{
nodes[usedNodes[i-1]].next = usedNodes[i];
}

//printing the nodes to console for verification
nodes[usedNodes[usedNodes.length - 1]].next = NUL;

for (int i = 0; i <= numStudents; i++)
{
if (nodes[i].student != null)
{
System.out.println(nodes[i].toString());
}
}
}

最终编辑:这是我完整的功能类,其中包含的方法未发布在我的原始代码中(我包含了删除、替换和检索学生成绩的其他方法,希望它将来可以帮助其他人) 。现在,它按照应有的方式排序(降序),并且我不是在插入时排序,而是在调用 printData() 方法期间对列表进行排序。我非常感谢 Andreas 帮助我解决了这个任务,没有你我不可能完成这个任务! (至少在需要的指定时间范围内):

// Returns the index of the next available node from the free list
// and updates the free list index
protected int getNode()
{
hold = free;
free = nodes[free].next;
return hold;
}

//Frees the node at array position index by linking it into the
//free list
protected void freeNode(int index)
{
nodes[index].next = free;
free = index;
nodes[free].student = null;
}

private void find(int id)
{
found = false;
current = list;
previous = NUL;
while (!found || current != NUL )
{
if (nodes[current].student.getId() == id)
{
found = true;
location = current;
return;
}
previous = current;
current = nodes[current].next;
}
}

//Method used to add student grades to nodes array. After LL node
//is added, method sorts the used nodes in descending order based
//off the float value grade
public void addStudentGrade(Student stu, float grade)
{
if (isFull())
{
return;
}

int newNode = getNode();
nodes[newNode].student = stu;
nodes[newNode].grade = grade;
nodes[newNode].next = list;
list = newNode;
numStudents++;

//printing sorted data to console
printData();
}

public boolean removeStudentGrade(int id)
{
find(id);
if (found)
{
hold = location;
if (list == location)
{
list = nodes[list].next; // remove first node
System.out.println("Removing first node");
}
else
{
nodes[previous].next = nodes[location].next;
}
freeNode(hold);
numStudents--;
}
return found;
}

//updates student grade from nodes array based on student id
public void updateStudentGrade(int id, float newGrade)
{
find(id);
if (found)
{
hold = location;
nodes[hold].grade = newGrade;
}
else
{
System.out.println("Student ID not found. Please check data entry and try again.");
}
}

//Method that searches array for student by id #. if found,
//students grade is returned. if not, -1 is returned
public float findStudentGrade(int id)
{
find(id);
if (found)
{
hold = location;
return nodes[hold].grade;
}
else
{
System.out.println("Student ID not found. Please check data entry and try again.");
}
return -1;
}

//method that sorts student's grades for a specific course in descending order
private void sort()
{
//storing used nodes in temp Integer array which will be used
//to compare and sort the grade values of each node
Integer[] usedNodes = new Integer[numStudents];

for (int i = list, j = 0; i != NUL; i = nodes[i].next)
{
usedNodes[j++] = i;
}

//sorting the temp array in descending order using custom
//comparator and Arrays.sort from Java.util.Arrays.
Arrays.sort(usedNodes, (i1, i2) ->
{
return Float.compare(nodes[i2].grade, nodes[i1].grade);
});

//rebuilding the linked list after sorting
list = usedNodes[0];
for (int i = 1; i < usedNodes.length; i++)
{
nodes[usedNodes[i-1]].next = usedNodes[i];
}
nodes[usedNodes[usedNodes.length - 1]].next = NUL;
}
//Determines whether this list is full
public boolean isFull()
{
return (free == NUL);
}

//prints grades for a specific course. Calls sort()
//method before printing
public void printData()
{
current = list;
while (current != NUL)
{
sort();
System.out.println(nodes[current].toString());
previous = current;
current = nodes[current].next;
}
}

最佳答案

已更新原始答案假设您只是想在事后对列表进行排序,但仔细观察,您似乎想要插入排序,其中具有相同成绩的学生应按插入顺序排列。

为此,需要替换 addStudentGrade() 方法的大部分内容。您不是“排序”,而是搜索“插入点”,然后在其中添加新节点。

为了展示这一点,我将代码缩减为 MCVE (最小、完整和可验证的示例)。您可以在 IDEONE 上查看完整代码.

public void addStudentGrade(Object stu, float grade) {
// Search for insertion point (i.e. between prevIdx and nextIdx)
int prevIdx = NUL;
int nextIdx = list;
while (nextIdx != NUL && nodes[nextIdx].grade >= grade) {
prevIdx = nextIdx;
nextIdx = nodes[nextIdx].next;
}

// Initialize and insert new node
int newIdx = getNode();
nodes[newIdx].student = stu;
nodes[newIdx].grade = grade;
nodes[newIdx].next = nextIdx;
if (prevIdx == NUL)
list = newIdx;
else
nodes[prevIdx].next = newIdx;
}
<小时/>

原始答案

这是一个非常奇怪的类,它有一个基于索引的链表到一个固定大小的节点数组中。

我建议,要对非自由节点进行排序,首先将这些节点收集到数组中,或者更确切地说是它们的节点索引。由于您知道有多少个正在使用的节点,即 numStudents,因此创建一个该大小的新数组,然后用正在使用的节点的索引填充它。请注意,数组必须是 Integer[],而不是 int[],因此可以提供自定义比较器来进行排序:

Integer[] inuse = new Integer[numStudents];
for (int i = list, j = 0; i != NUL; i = nodes[i].next)
inuse[j++] = i;

然后使用自定义比较器对它们进行排序,例如使用 Java 8 lambda:

Arrays.sort(inuse, (i1, i2) -> {
return Float.compare(nodes[i2].grade, nodes[i1].grade); // descending
});

现在它们已经排序,您需要重新构建链表:

list = inuse[0];
for (int j = 1; j < inuse.length; j++)
nodes[inuse[j - 1]].next = inuse[j];
nodes[inuse[inuse.length - 1]].next = NUL;

我相信应该可以做到。

关于java - 如何对链表节点数组进行从高到低排序(基于float值),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42883046/

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