删除C++中的链表中间值？
让我们首先定义包含数据和指向下一个节点的指针的链表。
struct Node {
int data;
struct Node* next;
};接下来我们创建我们的createNode(intdata)函数，该函数将int数据作为参数，并在分配参数值后返回新创建的节点。指向该节点的下一个指针将为空。
Node* createNode(int data){
struct Node* newNode = new Node;
newNode->data = data;
newNode->next = NULL;
return newNode;
}现在我们有我们的deleteMiddle(structNode*head)函数，它接受根节点。如果根节点不为空，则它只是将中间值之前的节点的下一个值分配给中间值旁边的节点，并返回临时头，即修改后的头。
struct Node* deleteMiddle(struct Node* head){
if (head == NULL)
return NULL;
if (head->next == NULL) {
delete head;
return NULL;
}
Node* temphead = head;
int count = nodeCount(head);
int mid = count / 2;
while (mid-- > 1) {
head = head->next;
}
head->next = head->next->next;
return temphead;
}最后我们有我们的printList(Node*ptr)函数，它获取列表头并打印列表。
void printList(Node * ptr){
while (ptr!= NULL) {
cout << ptr->data << "->";
ptr = ptr->next;
}
cout << "NULL"<示例
让我们看看下面删除单链表中间的实现。
#include
using namespace std;
struct Node {
int data;
struct Node* next;
};
Node* createNode(int data){
struct Node* newNode = new Node;
newNode->data = data;
newNode->next = NULL;
return newNode;
}
int nodeCount(struct Node* head){
int count = 0;
while (head != NULL) {
head = head->next;
count++;
}
return count;
}
struct Node* deleteMiddle(struct Node* head){
if (head == NULL)
return NULL;
if (head->next == NULL) {
delete head;
return NULL;
}
Node* temphead = head;
int count = nodeCount(head);
int mid = count / 2;
while (mid-- > 1) {
head = head->next;
}
head->next = head->next->next;
return temphead;
}
void printList(Node * ptr){
while (ptr!= NULL) {
cout << ptr->data << "->";
ptr = ptr->next;
}
cout << "NULL"<next = createNode(4);
head->next->next = createNode(6);
head->next->next->next = createNode(8);
head->next->next->next->next = createNode(10);
cout << "Original linked list"<输出结果
上面的代码将产生以下输出-
Original linked list
2->4->6->8->10->NULL
After deleting the middle of the linked list
2->4->8->10->NULL