c++ listnode反转方法分享

在C++中实现链表节点的反转可以通过迭代或递归两种方法来实现。以下是其中一种方法的示例代码：

#include <iostream>struct ListNode {    int val;    ListNode* next;    ListNode(int x) : val(x), next(NULL) {}};ListNode* reverseList(ListNode* head) {    ListNode* prev = NULL;    ListNode* curr = head;    while (curr != NULL) {        ListNode* next = curr->next;        curr->next = prev;        prev = curr;        curr = next;    }    return prev;}void printList(ListNode* head) {    while (head != NULL) {        std::cout << head->val << " ";        head = head->next;    }    std::cout << std::endl;}int main() {    ListNode* head = new ListNode(1);    head->next = new ListNode(2);    head->next->next = new ListNode(3);    head->next->next->next = new ListNode(4);    std::cout << "Original list: ";    printList(head);    head = reverseList(head);    std::cout << "Reversed list: ";    printList(head);    return 0;}

该示例代码中定义了一个节点结构体ListNode，并实现了reverseList函数来反转链表。通过调用reverseList函数并打印结果，可以看到链表节点的顺序被成功反转。