c++ - 链表中的节点最后没有删除

Question

我的功能是假设删除我的 Linked_list 中的最后一个节点，每个节点都有一个项目编号和一个年龄。但是当我调用我的函数并显示我的 Linked_list 时，该函数仅将项目编号设置为零，并且假设删除时年龄保持不变。

void deleteLastNode(Record  * & head){
Record *temp=head;

while(temp->next!=NULL){

    temp=temp->next;
}
    delete temp;


return;

};

//my output                                    item number  item age
//before the calling the function: 1st node    5000         1
//                                2nd node     6753         8 

//after calling the delete function: 1st node  5000         1
//                                   2nd node     0         8


//desired output                               
//before the calling the function: 1st node    5000         1
//                                 2nd node    6753         8 

//after calling the delete function: 1st node  5000         1

Answer 1

我认为您应该沿着列表向下走，直到到达倒数第二个 节点，然后将该节点指向NULL 并删除最后一个节点。我还认为更改您的函数以返回对列表新头部的引用是有意义的。

Record* deleteLastNode(Record* & head) {
    Record* temp = head;

    // for an empty list, just return NULL
    if (head == NULL) return NULL;

    // for a list one just one element, delete the head, and return NULL
    if (temp->next == NULL) {
        delete temp;
        head = NULL;
        return head;
    }

    // otherwise walk down the list until reaching the second to last element
    while (temp->next->next != NULL) {
        temp = temp->next;
    }

    delete temp->next;
    temp->next = NULL;

    // return the head of list with its final node removed
    return head;
}