使用另一个列表对链表进行排序

Question

所以我试图对一个整数链表进行排序，但似乎找不到正确的方法，我的想法是采用无序列表，从中找到最大值，然后将其放入另一个列表.由于我不相信我可以从原始列表中删除节点而不将其进行双重链接，因此我计划将列表 1 中的节点设为零值，从而将其状态删除为最大值。因此，我打算运行它一定次数，每次都找到下一个最大值，直到列表 1 全为 0，并且列表 2 是列表 1 曾经的有序版本。我已经创建了一个函数来执行此操作，但它似乎不起作用，虽然我找不到问题。

函数

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

struct num_node *create(struct num_node *list, int x){
    struct num_node *current;

    if (list == NULL){
        list = (struct num_node*)malloc(sizeof(struct num_node));
        list->num = x;
        list->next = NULL;
        return(list);
    }
    else{
        current = (struct num_node *)malloc(sizeof(struct num_node));
        current->num = x;
        current->next = list;
        return(current);
    }
} 

void print_nums(struct num_node *list) {

    struct num_node *current;
    for (current = list; current != NULL; current = current->next)
        printf("%d\t", current->num);

}

struct num_node *sort_nums(struct num_node *list1, struct num_node *list2){
    struct num_node *current;
    struct num_node *large = list1;

    for (int i = 0; i < 25; i++){
        for (current = list1; current != NULL; current = current->next){
            if (current->num > large->num){
                large = current;
            }
        }
        create(list2, large->num);
        large->num = 0;
        return(list2);
    }
}

int sum(struct num_node *list){
    int total = 0;
    struct num_node *current;
    for (current = list; current != NULL; current = current->next){
        total = total + current->num;
    }

    return total;
}

float average(struct num_node *list){
    float total = 0;
    float count = 0;
    struct num_node *current;
    for (current = list; current != NULL; current = current->next){
        total = total + current->num;
        count++;
    }
    return total / count;
}

主要

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

#include "functions.h"

int main(){
    struct num_node *head = NULL;
    struct num_node *new_head = NULL;

    srand(time(NULL));

        for (int i = 1; i <= 25; i++){
            int x = rand() % 100;
            head = create(head, x);
        }

        print_nums(head);

        sort_nums(head, new_head);

        printf("\n");
        printf("\n");
        print_nums(new_head);

        printf("\n");
        printf("\n");
        printf("The total of all numbers is: ");
        printf("\t%d\n", sum(new_head));

        printf("The average of the numbers is: ");
        printf("\t%.3f\n", average(new_head));


}

Answer 1

回到最初的想法，示例代码通过查找最大节点对链表进行排序，然后将其从原始列表（原始列表的左侧）中删除并将该节点插入新列表的前面。请注意，面向合并的算法会快得多。

NODE * SortList(NODE * pList)
{
NODE * pNew = NULL;                     /* sorted list */
NODE **ppNode;                          /* ptr to ptr to node */
NODE **ppLargest;                       /* ptr to ptr to largest node */
NODE * pLargest;                        /* ptr to largest node */
    while(pList != NULL){               /* while list not empty */
        ppLargest = &pList;             /*  find largest node */
        ppNode = &((*ppLargest)->next);
        while(NULL != *ppNode){
            if((*ppNode)->data > (*ppLargest)->data)
                ppLargest = ppNode;
            ppNode = &((*ppNode)->next);
        }
        pLargest = *ppLargest;          /* move node to new */
        *ppLargest = pLargest->next;
        pLargest->next = pNew;
        pNew = pLargest;
    }    
    return(pNew);
}

Answer 2

你从sort_nums过早返回：

struct num_node *sort_nums(struct num_node *list1, struct num_node *list2){
    struct num_node *current;
    struct num_node *large = list1;

    for (int i = 0; i < 25; i++){
        for (current = list1; current != NULL; current = current->next){
            if (current->num > large->num){
                large = current;
            }
        }
        create(list2, large->num);
        large->num = 0;
        return(list2);   // This just adds the largest item to list2
    }
}

你需要：

struct num_node *sort_nums(struct num_node *list1, struct num_node *list2){
    struct num_node *current;
    struct num_node *large = list1;

    for (int i = 0; i < 25; i++){
        for (current = list1; current != NULL; current = current->next){
            if (current->num > large->num){
                large = current;
            }
        }
        list2 = create(list2, large->num);
        large->num = 0;
    }
    return(list2);
}

另外，您没有在 main 中使用 sort_nums 的返回值。你有：

sort_nums(head, new_head);

你需要：

new_head = sort_nums(head, new_head);

简化create

由于您将项目添加到create 的列表中，因此可以简化为：

struct num_node *create(struct num_node *list, int x){
   struct num_node *current = malloc(sizeof(struct num_node));
   current->num = x;
   current->next = list;
   return(current);
} 

简化sort_nums

您还可以简化sort_nums。你不需要第二个参数。您可以使用：

struct num_node *sort_nums(struct num_node *list1){
   struct num_node *list2 = NULL;
   struct num_node *current;
   struct num_node *large = list1;
   int i;

   for (i = 0; i < 25; i++){
      for (current = list1; current != NULL; current = current->next){
         if (current->num > large->num){
            large = current;
         }
      }
      list2 = create(list2, large->num);
      large->num = 0;
   }
   return(list2);
}

当然，你必须改变在 main 中使用它的方式。

new_head = sort_nums(head);