c - 链接功能不会按预期将最大项目放在列表的末尾

Question

我有一个链表，我必须确保无论链表的最大数目是多少，我都将它放在链表的末尾。我不确定我做错了什么。有人可以解释一下吗？

void moveAllMaxAtEnd(list A) {
    int max=0;
    link tmp=A->first;
    link curr=tmp->next;
    int i,count=0;
    while(curr!=NULL){          //This first while is where I find the max item
        if(curr->item>=max){
            max=curr->item;
            count++;
        }
        else{
            curr=curr->next;
        }
     }
     link prev=A->first;
     link curr1=prev->next;
     link tmp1;
     while(curr1->next!=NULL){   //In this loop I am trying to put the 
         if(curr1->item==max){   //max items at the end.
            prev->next=curr1->next;
            tmp1=prev->next;
            prev->next=curr1->next;
            curr1->next=tmp1;
         }
         else{
             prev=prev->next;
             curr1=curr1->next;
         }
     }

}

Answer 1

虽然我还没有测试过，但它应该可以工作。如果是作业，请不要看今天的答案(自己学习会更有帮助!)。

对您的列表元素进行一些绘制并查看链接行为。您会立即发现此方法的问题。

我的绘图示例

更正(请勿立即阅读)

void moveAllMaxAtEnd(list A) {
    int max=0;
    link tmp=A->first;
    link curr=tmp->next;
    int i,count=0;
    while(curr!=NULL){          //This first while is where I find the max item
        if(curr->item>=max){
            max=curr->item;
            count++; //not used, may be cleaned up
        }
        //else{  //else is not needed the code would be executed in next loop anyway
        curr=curr->next;
        //}
     }
     link prev=A->first;
     link curr1=prev->next;
     link tmp1;
     while(curr1->next!=NULL){   //In this loop I am trying to put the 
         if(curr1->item==max){   //max items at the end.
            prev->next=curr1->next;
            tmp1=curr1; // not prev->next;
            // prev->next=curr1->next; cur1 has not changed you re-do exactly the same
            // curr1->next=tmp1;
            curr1->next=NULL; //it will be the last element
         }
         else{
             prev=prev->next;
             curr1=curr1->next;
         }
     }
     curr1->next = tmp1; //add the element we cut before
}