C：链表的合并排序，排序函数中未正确捕获合并的子数组

Question

根据 Wikipedia 站点上的合并排序伪代码，我可以深入了解一个节点正确合并到两个节点列表中的基本情况。但是，我认为当它们返回递归链时，合并的部分会反转。

我正在使用 Makefile：

sort: main.c sort.c
    gcc -Wall -std=c99 -o $@ main.c sort.c ll.c

我正在使用包含头指针和尾指针的列表头结构。 Head 指向第一个节点，tail 指向最后一个节点。
合并功能正在工作，但这里是排序功能。

void list_sort(list_t *list)
{
printf("in sort 1\n");

    //base case of 0 or 1 element
    if (list->head == NULL || list->head->next == NULL) {
        return;
    }

    list_t *sublistA = list_create();
    assert(sublistA);
    list_t *sublistB = list_create();
    assert(sublistB);

    int len = Length(list);
    int mid = (len) / 2;
    printf("mid is %d\n", mid);
    int i = 0;
    element_t *current = list->head;
    //make sublists
    while (i < len) {
        if (i < mid) {  
            printf("append sublistA\n");
            list_append(sublistA, current->val);
        } else {
            printf("append sub B\n");
            list_append(sublistB, current->val);
        }
        i++;
        current = current->next;    
    }
    list_print(sublistA);
    list_print(sublistB);

    printf("going to sort A\n");
    list_sort(sublistA);

    printf("going to sort B\n");
    list_sort(sublistB);
    //this was just added to capture returned list from merge
    list_t* capture = NULL;
    assert(capture);//the assertion failed
    capture = merge(sublistA, sublistB);
    }

编辑： 断言失败，所以我切换到：

list_t* capture = list_create();
    assert(capture);

    capture = merge(sublistA, sublistB);

我确信它没有从 main、merge 和 sort 中的 printf 语句中正确捕获。

这是显示意外反转的命令行输出：

/* left and right lists to merge://at the start of the merge function
{ 8, }
{ 7, }
in merge while 1
in merge else 1
{ 7, }
final result
{ 7, 8, }//yay!  this is the last line of merge function
in merge//came right back again to merge
left and right lists to merge:
{ 9, }
{ 8, 7, }//doh!*/

编辑：这里是合并函数：

list_t *merge(list_t *left, list_t *right) {
    printf("in merge\n");
    list_t *result = list_create();
    assert(result);

    element_t *curr1 = left->head;
    element_t *curr2 = right->head;
    //list_t *result = list_create();
    printf("left and right lists to merge: \n");
    list_print(left);
    list_print(right);

    while (curr1 != NULL && curr2 != NULL) {
printf("in merge while 1\n");
        if (curr1->val <= curr2->val) {
            list_append(result, curr1->val);
            printf("merge while 1 if\n");
            curr1 = curr1->next;
            //curr2 = curr2->next;
        }
        else //curr1->val > curr2->val
        {
            printf("in merge else 1\n");
            list_append(result, curr2->val);
            curr2 = curr2->next;
        }
    }
    list_print(result);

    //leftovers need to be allocated
    while (curr1 != NULL) {
        list_append(result, curr1->val);
        curr1 = curr1->next;
    }

    while (curr2 != NULL) {
        list_append(result, curr2->val);
        curr2 = curr2->next;
    }

    //printf("curr1->val is %d\n", curr1->val);
    //printf("curr2->val is %d\n", curr2->val);
    list_destroy(left);
    list_destroy(right);
    printf("final result\n");
    list_print(result);

    return result;
}

Answer 1

我注意到您的 merge 函数通过破坏性地修改输入列表来工作：您将两个列表作为输入，生成一个新的合并列表作为输出，然后返回结果列表。但是，在您的 list_sort 函数中，您没有捕获返回的新列表，因此您没有更新提供的输入列表以保存新排序的序列。尝试从merge 捕获返回值并使用它来更新输入列表参数。