请问C++数据结构单链表处理数集交集并集为什么不能正常输出呢？代码如下谢谢各位！！！

设计要求是单链表处理交集并集，能自动对输入的重复数据进行处理。我是纯新手，求大神帮忙修改代码，希望能详细注释说明，最好帮我简化一下代码，谢谢！！！

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


typedef struct Node
{
    char data;
    Node* next;
}SeqList;

bool IsContain(SeqList List, int x);
bool IsContain1(SeqList* List, char x);
SeqList* Intersection(SeqList* List1, SeqList* List2);
SeqList* Union(SeqList* List1, SeqList* List2);
SeqList* CreateSeqList(char n);
void Print(SeqList* List);


//判断一个元素是否包含于一个集合
bool IsContain(SeqList* List, char x)
{
    SeqList* p = List->next;
    while (p)
    {
        if (p->data == x)
            return true;
        p = p->next;
    }
    return false;
}
//用于判断一个元素是否包含于一个不用头结点表示的集合
bool IsContain1(SeqList* List, char x)
{!

    SeqList* p = List;
    while (p != NULL)
    {
        if (p->data == x)
            return true;
        p = p->next;
    }
    return false;
}
//求交集
SeqList* Intersection(SeqList* List1, SeqList* List2)
{
    SeqList* Seq, *Node, *k;
    SeqList* p = List1->next;
    Seq = (SeqList*)malloc(sizeof(SeqList));
    k = Seq;
    while (p != NULL)
    {
        if (IsContain(List2, p->data))
        {
            Node = (SeqList*)malloc(sizeof(SeqList));
            Node->data = p->data;
            k->next = Node;
            k = k->next;
        }
        p = p->next;   /*向后遍历*/
    }
    return Seq;
}
//求并集
SeqList* Union(SeqList* List1, SeqList* List2)
{
    SeqList* Seq = (SeqList*)malloc(sizeof(SeqList));
    SeqList* p = List1->next;
    SeqList* q = List2->next;
    SeqList* k = Seq;
    while (p != NULL)
    {
        SeqList* node = (SeqList*)malloc(sizeof(SeqList));
        node->data = p->data;
        k->next = node;
        k = k->next;
        p = p->next;
    }
    while (q != NULL)
    {
        if (!IsContain(List1, q->data))
        {
            SeqList* node = (SeqList*)malloc(sizeof(SeqList));
            node->data = q->data;
            k->next = node;
            k = k->next;
        }
        q = q->next;
    }
    return Seq;
}
//创建一个集合
SeqList* CreateSeqList(char n)
{
    SeqList* Seq = (SeqList*)malloc(sizeof(SeqList));
    SeqList* Head = (SeqList*)malloc(sizeof(SeqList));
    Seq->data = '\n';
    Seq->next = NULL;
    // SeqList* p,*q;
    //p = Seq;
    //q = Seq;
    printf("输入集合%c的元素：\n", n);
    char a;
    do
    {
        scanf("%c", &a);
    } while (!((a >= 'a'&& a <= 'z') || (a >= '0'&&a <= '9') || (a >= 'A'&&a <= 'Z')) && a != '\n');
    while (a != '\n')
    {
        // q = Seq;
        if (IsContain1(Seq, a))
        {
            do{
                scanf("%c", &a);
            } while (!((a >= 'a'&&a <= 'z') || (a >= '0'&&a <= '9') || (a >= 'A'&&a <= 'Z')) && a != '\n');
        }
        else
        {
            SeqList* node = (SeqList*)malloc(sizeof(SeqList));
            node->data = a;
            node->next = Seq;
            Seq = node;
            do{
                scanf("%c", &a);
            } while (!((a >= 'a'&&a <= 'z') || (a >= '0'&&a <= '9') || (a >= 'A'&&a <= 'Z')) && a != '\n');
        }
    }
    Head->next = Seq;
    return Head;
}
//输出集合
void Print(SeqList* List)
{
    SeqList* p = List->next;
    printf("(");
    while (p->next != NULL)
    {
        printf("%c", p->data);
        p = p->next;
    }
    p = p->next;
    printf("%c", p->data);
    printf(")");
    printf("\n");
}




int main()
{

    SeqList*sq1, *sq2;
    char cname1, cname2;
    printf("输入第一个集合的名字:\n");
    scanf("%c", &cname1);
    getchar();
    sq1 = CreateSeqList(cname1);
    getchar();
    printf("输入第二个集合的名字:\n");
    scanf("%c", &cname2);
    getchar();
    sq2 = CreateSeqList(cname2);
    printf("集合%c：\n", cname1);
    Print(sq1);
    printf("集合%c：\n", cname2);
    Print(sq2);

    printf("交集\n");
    Print(Intersection(sq1, sq2));
    printf("并集\n");
    Print(Union(sq1, sq2));
    return 0;
}

多了一个感叹号

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


typedef struct Node
{
    char data;
    Node* next;
}SeqList;

bool IsContain(SeqList List, int x);
bool IsContain1(SeqList* List, char x);
SeqList* Intersection(SeqList* List1, SeqList* List2);
SeqList* Union(SeqList* List1, SeqList* List2);
SeqList* CreateSeqList(char n);
void Print(SeqList* List);


//判断一个元素是否包含于一个集合
bool IsContain(SeqList* List, char x)
{
    SeqList* p = List->next;
    while (p)
    {
        if (p->data == x)
            return true;
        p = p->next;
    }
    return false;
}
//用于判断一个元素是否包含于一个不用头结点表示的集合
bool IsContain1(SeqList* List, char x)
{
    SeqList* p = List;
    while (p != NULL)
    {
        if (p->data == x)
            return true;
        p = p->next;
    }
    return false;
}
//求交集
SeqList* Intersection(SeqList* List1, SeqList* List2)
{
    SeqList* Seq, *Node, *k;
    SeqList* p = List1->next;
    Seq = (SeqList*)malloc(sizeof(SeqList));
    k = Seq;
    while (p != NULL)
    {
        if (IsContain(List2, p->data))
        {
            Node = (SeqList*)malloc(sizeof(SeqList));
            Node->data = p->data;
            k->next = Node;
            k = k->next;
        }
        p = p->next;   /*向后遍历*/
    }
    return Seq;
}
//求并集
SeqList* Union(SeqList* List1, SeqList* List2)
{
    SeqList* Seq = (SeqList*)malloc(sizeof(SeqList));
    SeqList* p = List1->next;
    SeqList* q = List2->next;
    SeqList* k = Seq;
    while (p != NULL)
    {
        SeqList* node = (SeqList*)malloc(sizeof(SeqList));
        node->data = p->data;
        k->next = node;
        k = k->next;
        p = p->next;
    }
    while (q != NULL)
    {
        if (!IsContain(List1, q->data))
        {
            SeqList* node = (SeqList*)malloc(sizeof(SeqList));
            node->data = q->data;
            k->next = node;
            k = k->next;
        }
        q = q->next;
    }
    return Seq;
}
//创建一个集合
SeqList* CreateSeqList(char n)
{
    SeqList* Seq = (SeqList*)malloc(sizeof(SeqList));
    SeqList* Head = (SeqList*)malloc(sizeof(SeqList));
    Seq->data = '\n';
    Seq->next = NULL;
    // SeqList* p,*q;
    //p = Seq;
    //q = Seq;
    printf("输入集合%c的元素：\n", n);
    char a;
    do
    {
        scanf("%c", &a);
    } while (!((a >= 'a'&& a <= 'z') || (a >= '0'&&a <= '9') || (a >= 'A'&&a <= 'Z')) && a != '\n');
    while (a != '\n')
    {
        // q = Seq;
        if (IsContain1(Seq, a))
        {
            do{
                scanf("%c", &a);
            } while (!((a >= 'a'&&a <= 'z') || (a >= '0'&&a <= '9') || (a >= 'A'&&a <= 'Z')) && a != '\n');
        }
        else
        {
            SeqList* node = (SeqList*)malloc(sizeof(SeqList));
            node->data = a;
            node->next = Seq;
            Seq = node;
            do{
                scanf("%c", &a);
            } while (!((a >= 'a'&&a <= 'z') || (a >= '0'&&a <= '9') || (a >= 'A'&&a <= 'Z')) && a != '\n');
        }
    }
    Head->next = Seq;
    return Head;
}
//输出集合
void Print(SeqList* List)
{
    SeqList* p = List->next;
    printf("(");
    while (p->next != NULL)
    {
        printf("%c", p->data);
        p = p->next;
    }
    p = p->next;
    printf("%c", p->data);
    printf(")");
    printf("\n");
}




int main()
{

    SeqList*sq1, *sq2;
    char cname1, cname2;
    printf("输入第一个集合的名字:\n");
    scanf("%c", &cname1);
    getchar();
    sq1 = CreateSeqList(cname1);
    getchar();
    printf("输入第二个集合的名字:\n");
    scanf("%c", &cname2);
    getchar();
    sq2 = CreateSeqList(cname2);
    printf("集合%c：\n", cname1);
    Print(sq1);
    printf("集合%c：\n", cname2);
    Print(sq2);

    printf("交集\n");
    Print(Intersection(sq1, sq2));
    printf("并集\n");
    Print(Union(sq1, sq2));
    return 0;
}

bool IsContain1(SeqList* List, char x)
{!

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