两个链表合并且倒序输出C++代码

#include <iostream>
using namespace std;

struct MyNode {
    int val;
    MyNode *next;
};

MyNode* create_link(int arr[], int num) {
    MyNode *head, *p, *node;
    head = NULL;
    for(int i = 0; i <= num - 1; ++i) {
        p = new MyNode;
        p->val = arr[i];
        p->next = NULL;

        if(head == NULL) {
            head = p;
        } else {
            node->next = p;
        }
        node = p;
    }
    return head;
}

MyNode* merge_link(MyNode* head1, MyNode* head2) {
    if (!head1) {
        return head2;
    }
    if (!head2) {
        return head1;
    }

    if (head1->val < head2->val) {
        head1->next = merge_link(head1->next, head2);
        return head1;
    } else {
        head2->next = merge_link(head1, head2->next);
        return head2;
    }
}

MyNode* reverse_link(MyNode* head)
{
    MyNode* link_head = head;
    MyNode* link_tail;
    MyNode* preNode = NULL;
    MyNode* pNode = head;
    MyNode* afterNode = head->next;
    for (pNode = link_head; pNode != NULL; pNode = afterNode) {
        afterNode = pNode->next;
        pNode->next = preNode;
        preNode = pNode;
    }
    link_tail = preNode;
    link_head = link_tail;
    return link_head;
}

void print_link(MyNode* p) {
    while(p != NULL) {
        cout << p->val << " ";
        p = p->next;
    }
    cout << endl;
}

int main() {
    int arr1[] = {1, 5, 7};
    int arr2[] = {2, 4, 6};
    int num1 = sizeof(arr1) / sizeof(int);
    int num2 = sizeof(arr2) / sizeof(int);

    MyNode *head1, *head2, *new_head, *rev_head;
    head1 = create_link(arr1, num1);
    head2 = create_link(arr2, num2);
    print_link(head1);
    print_link(head2);

    cout << "merge_link: " << endl;
    new_head = merge_link(head1, head2);
    rev_head = reverse_link(new_head);
    print_link(rev_head);

    return 0;
}

运行结果

1 5 7 
2 4 6 
merge_link: 
7 6 5 4 2 1