c++解决1)实现链表的排序(升序) 2)实现两个有序链表的合并:A=AUB,要求合并后仍然有序。 提交前请将所有的提示信息去掉，只保留最后的输出结果。例如运行时:从键盘直接输入: 212 3123 输出结果为: 1 2 3 分别表示第一个链表元素个数为2，元素分别为1.2;第二个链表元素个数为3，元素分别为1.2,3.

好的，我可以帮你解决这两个问题。首先是链表的排序，可以使用冒泡排序或者快速排序等算法。以下是使用冒泡排序的示例代码：

#include <iostream>
using namespace std;

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

ListNode* sortList(ListNode* head) {
    ListNode *p, *q, *tail = NULL;
    do {
        p = head;
        q = p->next;
        head = q;
        while (q) {
            if (p->val > q->val) {
                p->next = q->next;
                q->next = p;
                if (head == q) head = p;
                if (tail == p) tail = q;
                p = q;
            } else {
                p = p->next;
            }
            q = p->next;
        }
        tail = p;
    } while (head != tail);
    return head;
}

int main() {
    int n, val;
    cin >> n;
    ListNode *head = NULL, *p = NULL;
    for (int i = 0; i < n; i++) {
        cin >> val;
        ListNode *node = new ListNode(val);
        if (p == NULL) {
            p = head = node;
        } else {
            p->next = node;
            p = node;
        }
    }
    head = sortList(head);
    while (head) {
        cout << head->val << " ";
        head = head->next;
    }
    return 0;
}

接下来是两个有序链表的合并，可以使用双指针法。以下是示例代码：

#include <iostream>
using namespace std;

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    ListNode *head = NULL, *p = NULL;
    while (l1 && l2) {
        ListNode *node;
        if (l1->val < l2->val) {
            node = new ListNode(l1->val);
            l1 = l1->next;
        } else {
            node = new ListNode(l2->val);
            l2 = l2->next;
        }
        if (p == NULL) {
            p = head = node;
        } else {
            p->next = node;
            p = node;
        }
    }
    if (l1) {
        if (p == NULL) {
            head = l1;
        } else {
            p->next = l1;
        }
    } else if (l2) {
        if (p == NULL) {
            head = l2;
        } else {
            p->next = l2;
        }
    }
    return head;
}

int main() {
    int n, m, val;
    cin >> n;
    ListNode *l1 = NULL, *p = NULL;
    for (int i = 0; i < n; i++) {
        cin >> val;
        ListNode *node = new ListNode(val);
        if (p == NULL) {
            p = l1 = node;
        } else {
            p->next = node;
            p = node;
        }
    }
    cin >> m;
    ListNode *l2 = NULL;
    p = NULL;
    for (int i = 0; i < m; i++) {
        cin >> val;
        ListNode *node = new ListNode(val);
        if (p == NULL) {
            p = l2 = node;
        } else {
            p->next = node;
            p = node;
        }
    }
    ListNode *head = mergeTwoLists(l1, l2);
    while (head) {
        cout << head->val << " ";
        head = head->next;
    }
    return 0;
}

希望能帮到你！