升序排序鏈表

Question

我無法根據其評分對鏈接列表進行排序。 我得到了三個任務，如果列表為空，則添加第一個節點，如果傳入的節點的等級小於第一個節點，則將其移到最前面。 如果它大於最后一個值，則將其向后推，否則按正確的順序放置節點。

我是積極的函數push（int r，string c），addFirst（int r，string c）和addAtFront（int r，string c）正常工作。 我在實現節點屬於最低值和最高值之間的情況時遇到了麻煩。

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);
        }
    }
    }

}

這是main中的循環作為測試：

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();
    }

我的輸出：

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

輸出應為：

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

我在實現這些中間節點而又不覆蓋列表中較大的值時遇到很多麻煩。 最后的情況使我發瘋，我嘗試了許多不同的事情。

Answer 1

我看到的是，每次創建新的SNode時，New_SNode-> next都不會分配給列表的其余部分。 每次打印列表時，不會顯示以前的SNode。

首先聲明一個SNode * tmp2;

您的for循環應為：

 tmp2 = tmp->next;
 tmp->next = new SNode(r,c);
 tmp->next->next->tmp2;

祝好運。