关于#链表#的问题，如何解决？

Question

调了一天程序了，题目是将两个链表合并病按照学号升降排序，不知道是哪里出了问题

Answer 1

实现的是2个有序链表按升序排，修改了合并处的代码

#include <stdio.h>
#include <stdlib.h>

struct circle{
    int ID;
    char name;
    struct circle* next;
};

int main() {
    struct circle* head = (struct circle*) malloc(sizeof(struct circle));
    head->next=NULL;
    struct circle* p;
    p = head;
    int n;
    printf("请输入表长：");
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        struct circle* s = (struct circle*) malloc(sizeof(struct circle));
        p->next = s;
        printf("请输入ID: ");
        scanf("%d", &(s->ID));
        printf("请输入name: ");
        getchar();
        scanf("%c", &(s->name));
        s->next = NULL;
        p = s;
    }
    p = head;

    struct circle* head1 = (struct circle*) malloc(sizeof(struct circle));
    head1->next=NULL;
    struct circle* p1;
    p1 = head1;
    int k;
    printf("请输入表长：");
    scanf("%d", &k);
    for (int i = 1; i <= k; i++) {
        struct circle* s = (struct circle*) malloc(sizeof(struct circle));
        p1->next = s;
        printf("请输入ID: ");
        scanf("%d", &(s->ID));
        printf("请输入name: ");
        getchar();
        scanf("%c", &(s->name));
        s->next = NULL;
        p1 = s;
    }
    p1 = head1;

    struct circle* head2 = (struct circle*) malloc(sizeof(struct circle));
    head2->next=NULL;
    struct circle* p2;
    p2 = head2;
    p = p->next;
    p1 = p1->next;

    while(p != NULL || p1 != NULL) {
        if (p1 == NULL || (p != NULL && p->ID < p1->ID)) {
            p2->next = p;
            p2 = p;
            p = p->next;
        } else {
            p2->next = p1;
            p2 = p1;
            p1 = p1->next;
        }
    }

    p2 = head2;
    printf("输出:");
    for (int i = 0; i < n+k; i++) {
        p2 = p2->next;
        printf("%d %c ", (p2->ID), (p2->name));
    }

}