如何用 C++ 删除链表中间部分?
首先,让我们定义我们的链表,其中包含数据和指向下一个节点的指针。
struct Node { int data; struct Node* next; };
接下来,我们创建 createNode(int data) 函数,它接受 int data 作为参数,并在分配参数值后返回新增的节点。指向节点的下一个指针将为 null。
Node* createNode(int data){ struct Node* newNode = new Node; newNode->data = data; newNode->next = NULL; return newNode; }
现在,我们有了 deleteMiddle(struct Node* head) 函数,它采用根节点。如果根节点不为 null,则它只是将中间值之前节点的下一个值分配给中间值之后的节点,并返回修改后的头 temphead。
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,我们有了 printList(Node *ptr) 函数,它采用链表头并打印这个链表。
void printList(Node * ptr){ while (ptr!= NULL) { cout << ptr->data << "->"; ptr = ptr->next; } cout << "NULL"<<endl; }
示例
让我们看一下以下删除单链表中间部分的实现。
#include <iostream> 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"<<endl; } int main(){ struct Node* head = createNode(2); head->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"<<endl; printList(head); head = deleteMiddle(head); cout<<endl; cout << "After deleting the middle of the linked list"<<endl; printList(head); return 0; }
输出
以上代码将生成以下输出 -
Original linked list 2->4->6->8->10->NULL After deleting the middle of the linked list 2->4->8->10->NULL
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