通过交换节点选择单个链接列表的类型（迭代方法？

Question

我试图通过交换节点本身来对单链列表执行选择排序，但是在所有输入之后，它看起来像我的sort（）函数无法正常工作。我想念什么或做错了什么，请有人帮我解决。

注意：所有函数和指针的名称几乎都说明了他们的任务，所以我认为不添加注释，但是如果有人想要注释，请告诉我。

注意：在swap_node(int i, int j)中，i和j是需要交换的两个节点的位置。

#include <stdio.h>
#include <stdlib.h>

typedef struct node
{
    int data;
    struct node *next;
} node;

node *head = NULL, *prev = NULL, *next = NULL;

static int k = 0;//Global variable K which will store the no of nodes created so far.

node *make_node()
{
    k++;
    return ((node *)malloc(sizeof(node)));
}

void push()
{
    if (k == 0)
    {
        next = make_node();
        printf("Enter Data:");
        scanf("%d", &next->data);
        next->next = NULL;
        head = next;
        prev = next;
    }
    else
    {
        next = make_node();
        printf("Enter Data:");
        scanf("%d", &next->data);
        next->next = NULL;
        prev->next = next;
        prev = next;
    }
}

node *x = NULL, *y = NULL;

void swap_node(int i, int j)
{
    node *prevX = NULL, *currX = head, *prevY = NULL, *currY = head;
    if (i == 1)
    {
        for (int l = 1; l < j; l++)
        {
            prevY = currY;
            currY = currY->next;
        }
        head = currY;
        if (j - i == 1)
        {
            currX->next = currY->next;
            currY->next = currX;
        }
        else
        {
            node *temp = currX->next;
            currX->next = currY->next;
            currY->next = temp;
            prevY->next = currX;
        }
    }
    else
    {
        for (int l = 1; l < i; l++)
        {
            prevX = currX;
            currX = currX->next;
        }
        for (int l = 1; l < j; l++)
        {
            prevY = currY;
            currY = currY->next;
        }
        if (j - i == 1)
        {
            prevX->next = currY;
            currX->next = currY->next;
            currY->next = currX;
        }
        else
        {
            node *temp = currX->next;
            prevX->next = currY;
            prevY->next = currX;
            currX->next = currY->next;
            currY->next = temp;
        }
    }
    x = currX;
    y = currY;
}

void sort()
{
    x = head;
    int i = 0, j = 0;
    for (i = 1; i < k; i++)
    {
        y = x->next;
        for (j = i + 1; j <= k; j++)
        {
            if (x->data > y->data)
            {
                swap_node(i, j);
            }
            y = y->next;
        }
        x = x->next;
    }
}

void print_node()
{
    printf("------------Printing Node--------------\n");
    node *temp = head;
    do
    {
        printf("%d\n", temp->data);
        temp = temp->next;
    } while (temp != NULL);
}

void main(void)
{
    int choice;
    printf("MENU\n1-PUSH\n2-Print node\n");
    do
    {
        printf("Enter Your Choice:");
        scanf("%d", &choice);
        switch (choice)
        {
        case 1:
            push();
            break;
        case 2:
            sort();
            print_node();
            break;
        default:
            printf("Wrong Choice!");
        }

    } while (choice == 1);
}

Answer 1

为什么不只是交换内容：

void swap_content (node *first, node *second) {
    int med = first->data; 
    first->data = second->data;
    second->data = med;
}