我无法弄明白，为什么我的代码插入到排序的双向链表中，在某些测试用例中失败。请让我知道。我不知道测试用例，它们是系统生成的。
Node* SortedInsert(Node *head,int data)
{
// Complete this function
// Do not write the main method.
Node * temp = (Node*)malloc(sizeof(Node));
temp->data = data;
temp->next = NULL;
temp->prev = NULL;
if (head == NULL)
{
head = temp;
return head;
}
if (temp->data <= head->data)
{
temp->next = head;
head->prev = temp;
head = temp;
return head;
}
Node *curr = head;
while (curr->next!=NULL)
{
if (temp->data <= curr->data)
{
curr->prev->next = temp;
temp->prev = curr->prev;
temp->next = curr;
curr->prev = temp;
return head;
}
curr = curr->next;
}
curr->next = temp;
temp->prev = curr;
return head;
}
谢谢
答案 0 :(得分：7)
到达最后一个节点后，您应该再次将其数据与新节点进行比较并相应插入。
curr->next = temp;
temp->prev = curr;
return head;
}
如果执行到达此部分，则当前curr指向最后一个节点。现在你应该再次比较它的数据如下：
if (temp->data <= curr->data)
{ // insert before last node
curr->prev->next = temp;
temp->prev = curr->prev;
temp->next = curr;
curr->prev = temp;
return head;
}
// else insert at the end.
curr->next = temp;
temp->prev = curr;
return head;
}
答案 1 :(得分：1)
或者，您可以在列表末尾为节点编写if条件
Node* SortedInsert(Node *head,int data)
{
struct Node* p = head;
struct Node* q = NULL;
struct Node* r = new Node;
r->data=data;
r->prev=NULL;
r->next=NULL;
if(p==NULL){
p=r;
head=p;
}
else if(p!=NULL&&p->data>r->data){
p->prev=r;
r->next=p;
p->next=NULL;
head = r;
}
else{
p=head;
while(p!=NULL) {
if(p!=NULL&&p->data>r->data){ //If node is not at the end of list and smaller than some node
(p->prev)->next = r;
r->next = p;
r->prev = p->prev;
p->prev=r;
return head;
}
else if(p->next==NULL) //If node is at the end of list
{
p->next = r;
r->prev = p;
r->next = NULL;
return head;
}
else{}
p=p->next;
}
}
return head;
}
答案 2 :(得分：0)
/*
* For reference:
*
* link to the problem :-
* https://www.hackerrank.com/challenges/insert-a-node-into-a-sorted-doubly-linked-list
*
* DoublyLinkedListNode {
* int data;
* DoublyLinkedListNode* next;
* DoublyLinkedListNode* prev;
* };
*
* Function to insert a node in sorted DLL with given data & return head
*/
DoublyLinkedListNode* sortedInsert(DoublyLinkedListNode* head, int data) {
DoublyLinkedListNode** prev = &head;
DoublyLinkedListNode* newNode = new DoublyLinkedListNode(data);
while ( *prev && ((*prev)->data <= data) ) {
prev = &(*prev)->next;
}
newNode->prev = *prev;
newNode->next = *prev;
*prev = newNode;
return head;
}
此解决方案在所有情况下均有效（在中间插入，在头部插入，在末端插入等）。我只是试图在没有任何条件的情况下编写代码。在搜索某些错误时，我最终遇到了这个SO线程。如果此解决方案不适用于某些测试用例，请纠正我。
答案 3 :(得分：0)
这是我通过所有测试用例的代码。
DoublyLinkedListNode* sortedInsert(DoublyLinkedListNode* head, int data) {
DoublyLinkedListNode *temp = head;
if (!temp) {
temp = new DoublyLinkedListNode(data);
head = temp;
return head;
}
while (temp->data < data && temp->next != nullptr) {
temp = temp->next;
}
if (temp->data > data) {
DoublyLinkedListNode *t = new DoublyLinkedListNode(data);
t->next = temp;
if (temp == head) {
head = t;
temp->prev = t;
return head;
}
temp->prev->next = t;
t->next = temp;
temp->prev = t;
} else {
DoublyLinkedListNode *t = new DoublyLinkedListNode(data);
t->next = temp->next;
temp->next = t;
temp->next->prev = temp;
if (t->next != nullptr) t->next->prev = t;
}
return head;
}
答案 4 :(得分：-1)
以下是一些代码，以便在排序的双向链接列表
中插入节点Node* SortedInsert(Node *head,int data)
{
struct Node *temp;
struct Node *newNode = new Node();
newNode->data=data;
newNode->next=NULL;
newNode->prev=NULL;
temp=head;
if(head==NULL)
{
head = newNode;
}
else
{
while(temp!=NULL)
{
if(temp->data<data&&temp->next!=NULL)
{
temp=temp->next;
}
else
{
if(temp->next==NULL&&temp->data<data)
{
temp->next = newNode;
newNode->prev = temp;
break;
}
else if(temp->prev==NULL&&temp->data>data)
{
newNode->next=temp;
temp->prev = newNode;
head=newNode;
break;
}
else
{
newNode->next = temp;
newNode->prev = temp->prev;
temp->prev->next =newNode;
temp->prev=newNode;
break;
}
}
}
}
return head;
}
答案 5 :(得分：-1)
这是一个完整的C ++程序，用于在排序的双向链表中的适当位置插入节点：
void InsertInSortedDoublyLinkedList()
{
Node *head;
head = CreateDoublyLinkList(3);
PrintDoublyLinkedList(head);
head = SortedInsert(head, 6);
printf("\n Linked list after insertion in sorted order\n");
PrintDoublyLinkedList(head);
}
void PrintDoublyLinkedList(Node *head)
{
Node *temp = head;
printf("NULL -> ");
while (temp)
{
printf("%d ->", temp->data);
temp = temp->next;
}
printf("NULL");
}
Node* CreateDoublyLinkList(int numberOfNodes)
{
Node *head = NULL, *temp, *temp1;
int startingValue = 3;
if (numberOfNodes == 0)
{
return head;
}
temp = (Node*)malloc(sizeof(Node));
temp->data = startingValue;
temp->next = NULL;
temp->prev = NULL;
head = temp;
startingValue += 2;
numberOfNodes--;
for (; numberOfNodes > 0; numberOfNodes--, startingValue += 2, temp = temp->next)
{
temp1 = (Node*)malloc(sizeof(Node));
temp1->data = startingValue;
temp1->next = NULL;
temp1->prev = temp;
temp->next = temp1;
}
return head;
}
Node* SortedInsert(Node *head,int data)
{
Node *temp = NULL, *temp1,*newNode;
int nodeInserted = 0;
if (head == NULL)
{
head = (Node*) malloc(sizeof(Node));
head->data = data;
head->next = NULL;
head->prev = NULL;
}
else
{
if (head->data > data)
{
//insertion need to take place before head itself.
newNode = (Node*)malloc(sizeof(Node));
newNode->data = data;
newNode->next = head;
newNode->prev = NULL;
head->prev = newNode;
head = newNode;
}
else
{
temp1 = head;
temp = head ->next;
while (temp)
{
if (temp->data > data)
{
//we need to insert the node before temp
newNode = (Node*)malloc(sizeof(Node));
newNode->data = data;
newNode->prev = temp1;
newNode->next = temp;
temp1->next = newNode;
temp->prev = newNode;
nodeInserted = 1;
break;
}
temp1 = temp;
temp = temp->next;
}
if (!nodeInserted)
{
//node insertion need to take place at tail.
newNode = (Node*)malloc(sizeof(Node));
newNode->data = data;
newNode->prev = temp1;
newNode->next = NULL;
temp1->next = newNode;
}
}
}
return head;
}
答案 6 :(得分：-2)
struct Node
{
int data;
Node *next;
Node *prev;
}
Node* SortedInsert(Node *head,int data)
{
Node* p1,*p2;
int n=10;
p2=(Node*)malloc(sizeof(struct Node));
p2->next=NULL;
p2->prev=NULL;
p2->data=data;
p1=head;
if(p1==NULL)
return p2;
while(p1->next!=NULL)
{
if(p1->data<data)
p1=p1->next;
else
break;
}
/*Three cases arise when p1->next == NUll i.e. we are end of list
case 1: insert after the end
case 2: insert in between last and second last node
case 3: insert node at the beginning of the list
*/
if(p1->next==NULL)
{
if(p1->data<data)
{
p2->prev=p1;
p1->next=p2;
}
else if(p1->data > data && p1->prev!=NULL)
{
p2->next=p1;
p2->prev=p1->prev;
if(p1->prev!=NULL)
p1->prev->next=p2;
p1->prev=p2;
}
else
{
p2->next=p1;
p2->prev=p1->prev;
if(p1->prev!=NULL)
p1->prev->next=p2;
p1->prev=p2;
head=p2;
}
}
// Here we have only one case where new node is inserted between two nodes of the list
else
{
p2->next=p1;
p2->prev=p1->prev;
if(p1->prev!=NULL)
p1->prev->next=p2;
p1->prev=p2;
}
return head;
}