C-链表-气泡排序算法中的交换节点

问题描述 投票:0回答:1

:我想避免向StackOverflow社区寻求帮助,因为我知道并支持这种问题不会因案例而异并且多次seen and reseen会增加任何价值。那就是说可以随意投票,但是我非常需要帮助:是我正在尝试的5小时,明天我要考试!

C中,我必须编写一个BubbleSort函数,该函数交换节点而不是值] LinkedList,但是我无法完成它。这是代码(如您所见,顺序不正确):

#include <stdio.h>
#include <malloc.h> // malloc, free
#include <stddef.h> // NULL

// defining 'int' as data_t
typedef int data_t;

typedef struct node_s {
    data_t data;
    struct node_s* next;
} node_t;

node_t** list_new() {
    node_t** list = malloc(sizeof **list);
    if (!list) return NULL;
    *list = NULL;
    return list;
}

void list_delete(node_t** list) {
    node_t *node, *next;
    for (node = *list; node; node = next) {
        next = node->next;
        free(node);
    }
}

node_t* list_push_front(node_t** list, data_t data) {
    node_t* node = malloc(sizeof(node_t));
    if (!node) return NULL;
    node->data = data;
    node->next = *list;
    return *list = node;
}

// IS EASY TO SWAP THE VALUES
/*
void swap(data_t* a, data_t* b) {
    data_t c = *a;
    *a = *b;
    *b = c;
}

void simple_bubble_sort(node_t** list) {
    for(node_t* i = *list; i; i = i->next)
        for(node_t* j = *list; j->next; j = j->next)
            if(j->data > j->next->data)
                swap(&(j->data), &(j->next->data));

}
*/
// MUCH LESS EASY TO SWAP THE NODES
void swap_node(node_t** prev_node_A, node_t** node_A, node_t** node_B) {
        node_t *last_node = (*node_B)->next;
        node_t *first_node = *node_A;
        node_t *second_node = *node_B;
        if (prev_node_A) {
            (*prev_node_A)->next = second_node;
            (*prev_node_A)->next->next = first_node;
            (*prev_node_A)->next->next->next = last_node;
        } else {
            (*node_A) = second_node;
            (*node_A)->next = first_node;
            (*node_A)->next->next = last_node;
        }
}

void simple_bubble_sort(node_t** list) {

    for(node_t* i = *list; i->next; i = i->next)
        for (node_t *j = *list; j->next->next; j = j->next) {
            if (j == *list) {
                if (j->data > j->next->data) {
                    *list = j->next;
                    swap_node(NULL, &j, &(j->next));
                }
            }
            else {
                if (j->next->data > j->next->next->data)
                    swap_node(&j, &(j->next), &(j->next->next));
            }
            //printf("%i,%i | %i, %i, %i, %i \n", i->data, j->data, (*list)->data, (*list)->next->data, (*list)->next->next->data, (*list)->next->next->next->data);
            //system("pause");
        }
}


int main() {
    // Create List
    node_t** list = list_new();
    if (!list)
        goto memory_allocation_failure;

    // Add values to List
    for(int i=0; i<10; i++)
        if (!list_push_front(list, i))
            goto memory_allocation_failure;

    // Print List
    for (node_t* node = *list; node != NULL; node = node->next)
        printf("%i\n", node->data);

    // Swap Test
    //swap_node(NULL, &(*list), &((*list)->next));
    //swap_node(&(*list), &((*list)->next), &((*list)->next->next));

    // Sort List
    printf("-- Bubble Sort --\n");
    simple_bubble_sort(list);

    // Print List
    for (node_t* node = *list; node != NULL; node = node->next)
        printf("%i\n", node->data);

    // Delete List
    list_delete(list);
    return 0;

    // Error Handler
    memory_allocation_failure:
        printf("Memory Allocation Failure");
        return 1;
}

谢谢,马可。

c linked-list swap bubble-sort
1个回答
0
投票
这里是功能swap

void swap( node_t **current ) { node_t *tmp = ( *current )->next->next; ( *current )->next->next = *current; *current = ( *current )->next; ( *current )->next->next = tmp; }

这是函数simple_bubble_sort

void simple_bubble_sort( node_t **head ) { if ( *head ) { for ( node_t **first = head, *sorted = NULL, *last = sorted; ( *first )->next != last; last = sorted ) { sorted = ( *first )->next; for ( node_t **current = first; ( *current )->next != last; current = &( *current )->next ) { if ( ( *current )->next->data < ( *current )->data ) { swap( current ); sorted = ( *current )->next; } } } } }

调查他们。
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