我无法根据其评分对链接列表进行排序。我得到了三个任务，如果列表为空，则添加第一个节点，如果传入的节点的等级小于第一个节点，则将其移到最前面。如果它大于最后一个值，则将其向后推，否则按正确的顺序放置节点。
我是积极的函数push(int r，string c)，addFirst(int r，string c)和addAtFront(int r，string c)正常工作。我在实现节点属于最低值和最高值之间的情况时遇到了麻烦。
排序功能如下:
void SLL::insertInOrder(int r, string c){
SNode *tmp = new SNode(r,c);
if(first == NULL){
addFirst(tmp->rating,tmp->comments);
}
else if(tmp->rating < first->rating){
addAtFront(r,c);
}
else if(tmp->rating > last->rating){
push(r,c);
}
else{
for(tmp =first; tmp->next != NULL; tmp = tmp->next){
if(tmp->rating < tmp->next->rating){
tmp->next = new SNode(r,c);
}
}
}
}
这是main中的循环作为测试:
int r[10] = {9,8,4,5,11,10,3,6,8,2};
string s[10] = {"really good!","loved it","mediocre",
"okay, not great","best book ever!", "awesome!",
"boring","not bad","definitely worth reading", "terrible!"};
SLL *list = new SLL();
for (int i = 0; i < 10; i++){
list->insertInOrder(r[i],s[i]);
list->printSLL();
}
我的输出:
Rating: 9,Comments: really good!
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 4,Comments: mediocre
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 4,Comments: mediocre
Rating: 5,Comments: okay, not great
Rating: 4,Comments: mediocre
Rating: 5,Comments: okay, not great
Rating: 4,Comments: mediocre
Rating: 10,Comments: awesome!
Rating: 3,Comments: boring
Rating: 4,Comments: mediocre
Rating: 10,Comments: awesome!
Rating: 3,Comments: boring
Rating: 6,Comments: not bad
Rating: 3,Comments: boring
Rating: 8,Comments: definitely worth reading
Rating: 2,Comments: terrible!
Rating: 3,Comments: boring
Rating: 8,Comments: definitely worth reading
输出应为:
Rating: 9,Comments: really good!
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 4,Comments: mediocre
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 4,Comments: mediocre
Rating: 5,Comments: okay, not great
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 4,Comments: mediocre
Rating: 5,Comments: okay, not great
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 11,Comments: best book ever!
Rating: 4,Comments: mediocre
Rating: 5,Comments: okay, not great
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 10,Comments: awesome!
Rating: 11,Comments: best book ever!
Rating: 3,Comments: boring
Rating: 4,Comments: mediocre
Rating: 5,Comments: okay, not great
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 10,Comments: awesome!
Rating: 11,Comments: best book ever!
Rating: 3,Comments: boring
Rating: 4,Comments: mediocre
Rating: 5,Comments: okay, not great
Rating: 6,Comments: not bad
Rating: 8,Comments: loved it
Rating: 9,Comments: really good!
Rating: 10,Comments: awesome!
Rating: 11,Comments: best book ever
我在不覆盖列表中较大的值的情况下实现这些中间节点有很多麻烦。最后的情况使我发疯，我尝试了许多不同的事情。
最佳答案
我看到的是，每次创建新的SNode时，New_SNode-> next都不会分配给列表的其余部分。
每次打印列表时，不会显示以前的SNode。
在开始时声明一个
SNode * tmp2;
您的for循环应为:
tmp2 = tmp->next;
tmp->next = new SNode(r,c);
tmp->next->next->tmp2;
祝好运。
关于c++ - 升序排序链表，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/52675301/