Question

我试图对双重链表进行排序，但我遇到了一些麻烦。我是C中的菜鸟，我想我的问题是指针......

我无法看到如何在列表中交换两个位置，所以也许这就是问题所在。

我试图通过使用Bubblesort对它进行排序，即使知道它的复杂性不是很好，因为我仍在学习，认为这是一种简单的开始方式。

我也试过阅读一些关于在链表中交换元素以及如何对它们进行排序的事情，但我真的遇到了这个问题...

PS：我用m-＆gt;开始了for;下一次因为我的列表有一个标题（m）。

PS2：我收到错误＆＃34;要求成员'下一个'不是结构或工会＆＃34;，并且不知道如何解决它

struct segment {
   int x, y;   /// position
   char c;   // letter 
   struct segment* next;
   struct segment* prev; 
};


void sortingSegments(struct segment* m) {
   struct segment **j; struct segment **i;

   for(i = &((m->next)->next); i !=NULL; i = i->next) {
          for(j = &(m->next); j == i; j = j->next) {
              if ((*j)->c > (*i)->c) {
                  struct segment **aux;
                  aux = i;
                  (*aux)->next = (*i)->next;
                  (*aux)->prev = (*i)->prev;

                  i = j;
                  (*i)->next = (*j)->next;
                  (*i)->prev = (*j)->prev;

                  j = aux;
                  (*j)->prev = (*aux)->prev;
                  (*j)->next = (*aux)->next;
              }
          }
   } 
}

Answer 1

请阅读评论并尝试了解节点的链接。

它基于simple bubble sort described in wikipedia。

void sortingSegments(struct segment** m) {
    struct segment *i, *tmp, *prev, *next;
    int swapped = 1;
    /*
     * https://en.wikipedia.org/wiki/Bubble_sort#Pseudocode_implementation
     *  n = length(A)
     *  repeat
     *    swapped = false
     *    for i = 1 to n-1 inclusive do
     *      //if this pair is out of order
     *      if A[i - 1] > A[i] then
     *        // swap them and remember something changed
     *        swap(A[i - 1], A[i])
     *        swapped = true
     *      end if
     *    end for
     *  until not swapped
     */

     // looping until no more swaps left
     while (swapped) { 
        swapped = 0;
        // we begin from the second item at each iteration
        for (i = (*m)->next; i; i = i->next) { 
            // we need to swap i with i->prev
            if (i->prev->c > i->c) { 
                prev = i->prev;
                next = i->next;
                // swapping first and second elements,
                // so update m to reflect the change made 
                // to the head of the list
                if (prev == *m) { 
                    *m = i;
                } 
                // so there is a prev of prev, update that two links
                else { 
                    prev->prev->next = i;
                    i->prev = prev->prev;
                }
                // so there is a next, update that two links
                if (next) { 
                    next->prev = prev;
                    prev->next = next;
                }
                // no next element, mark the end of the list
                else { 
                    prev->next = NULL;
                }
                // this is obvious, prev now becomes i's next
                prev->prev = i; 
                i->next = prev;

                // this is needed to reflect the change in i
                i = prev; 
                swapped = 1;
            }
        }
    }
}

链接列表中的C Bubblesort

1 个答案: