Question

为什么排序链接列表的冒泡排序实现不起作用？

它根本没有排序。代码的主要问题区域是sort函数。我已经验证了在调用sort函数之前，主要工作正常并且链表的最后一个指针（下一个）被设置为null。链接列表的链应该在字符串比较块内的if语句中链接，并且应该返回列表中的第一个链接，while语句将调用该链接以访问已经整理好的所有成员但是，它似乎没有起作用。

#include <stdbool.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>

#define NAME_SIZE 20

typedef struct record rec;
struct record
{
    char name[NAME_SIZE];
    int age;
    rec* next;
};

void getname(char* name);
rec* sort(rec**First,int i);/*we need to change the addresses they refer to */

int main(void)
{
    rec* Current = NULL;
    rec* Previous = NULL;
    rec* First = NULL;
    char check = '\0';
    int i = 0;

    for(; ;)
    {
        fflush(stdin);
        printf("\nDo you want to enter a%s record?(y/n): ", (check=='\0')?"":"nother");
        scanf("%c", &check);
        if(check == 'n')
            break;

        Current = (rec*)malloc(sizeof(rec));
        if(First == NULL)
        {
            First = Current;
        }
        if(Previous != NULL)
        {
            Previous->next = Current;
        }
        printf("\n");
        printf("\nPlease enter your name: ");
        getname(Current->name);
        printf("\nPlease enter your age: ");
        scanf("%d", &Current->age);
        Previous = Current;
        Current->next = NULL;
        i++;
    }

    Current = sort(&First, i);


    while(Current != NULL)
    {
        printf("\n%s is %d years old ", Current->name, Current->age);
        Current = Current->next;
    }

    return 0;
}

void getname(char *name)
{
    fflush(stdin);
    fgets(name, NAME_SIZE, stdin);
    int length = strlen(name);
    if(name[length - 1] == '\n')
        name[length -1 ] = '\0';
    return;
}


rec* sort(rec** first, int numbers)
{
    rec* pTemp1 = NULL;
    rec* pTemp2 = NULL;
    rec* Temp_first = *first;

    for(int j = 0; j < numbers; j++)
    {
        pTemp1 = *first;
        if(((*first) = (*first)->next)==NULL)
        {/*if end is reached, then break;*/
            break;
        }

        if(strcmp((*first)->name, ((*first)->next->name)) > 0)
        {


            if(((*first)->next) != NULL)
            {
                printf("\n***XXX***Entered***XXX");
                pTemp2 = (*first)->next;
                (*first)->next = pTemp1;
                pTemp1->next = pTemp2;
            }
        }

    }

    return Temp_first;

}

（省略了释放内存以简明扼要）。输入

atest  
aatest  
btest  
abtest

并且输出没有排序：

atest  
aatest  
abtest

Answer 1

我认为你在bubble sort函数中做错了，你似乎正在重复列表的相邻比较N次，这没关系，但你只是对前两个元素进行比较的清单。回想一下，冒泡排序是一种O（N ^ 2）算法。您需要将相邻元素检查放入另一个循环中。

此外，您可以通过更改外部循环来优化它，以依赖于冒泡排序变体，该变量继续传递，直到列表中没有反转为止。我根据自己的风格更改了冒泡排序方法，我能够看到所需的输出：

rec* sort(rec** first, int numbers)
{
    int bSwap = 0;
    if(*first == NULL || (*first)->next == NULL)
      return;

    // outer loop for iterating till list is sorted
    while(!bSwap) {
      // iterating over the list comparing and
      // swapping adjacent elements
      rec *curNode = *first;
      rec *prev = *first;
      rec *fwd = (*first)->next;
      bSwap = 1;

      while(fwd) {
        int cmp = strcmp(curNode->name, fwd->name);

        if(cmp > 0) { // inversion found, swap and proceed forward
          if(curNode == *first)
            *first = fwd;
          else
            prev->next = fwd;

          curNode->next = fwd->next;
          fwd->next = curNode;
          prev = fwd;
          fwd = curNode->next;
          bSwap = 0;
        }
        else { // proceed forward
          curNode = prev->next;
          prev = fwd;
          fwd = fwd->next;
        }
      }

    }
    return *first;
}

这是我运行时得到的结果：

~/Documents/src : $ ./a.out 

Do you want to enter a record?(y/n): y


Please enter your name: Rohan

Please enter your age: 22

Do you want to enter another record?(y/n): y


Please enter your name: Hema

Please enter your age: 20

Do you want to enter another record?(y/n): y


Please enter your name: Asanala

Please enter your age: 31

Do you want to enter another record?(y/n): n

Asanala is 31 years old 
Hema is 20 years old 
Rohan is 22 years old

我希望这会有所帮助。

用冒泡排序对链表进行排序

1 个答案: