8---------单链表算法的简单应用和巩固

#include <stdio.h>
#include <stdlib.h>
/**带头结点的单链表
* 链表的逆置
* 两个链表的交集合问题(链表初始无序)
* 两个链表的并集问题(链表初始无序)
* 链表A-链表B差集问题(链表初始无序)
* 找到单链表倒数第k个结点并打印
*/

//链表结点结构体定义
typedef struct Node{
int data;
struct Node * next;
}Node;

//链表的逆置
void inverse(Node * node){
Node * q,* p;
q = node -> next;
while(q){
p = q->next;
q->next = node->next;
node -> next = q;
q = p;
}
}

//链表的交集合问题,AnB=C,不破坏原本A和B链表结构
void itersection(Node * A,Node * B,Node * C){
Node * q = A->next;
Node * p = B->next;
Node * c = C;   //c本身是带头结点的单链表
Node * s;
while(q){
while(p){
if(p->data == q->data){
s = (Node *)malloc(sizeof(Node));
s -> data = q->data;
s -> next = NULL;
c ->next = s;
c = c->next;
}
p = p->next;
}
q = q->next;
}
}

//两个链表的并集问题，AuB = C,思路，先将一个链表的所有数据线存到C，然后再B链表中的结点一一于C中的结点比较，如果有相同的就不加入
void combine(Node * A,Node * B,Node * C){
Node * q = A->next;
Node * p = B->next;
Node * justc;
Node * c = C;   //C本身是带头结点
Node * s;
int flag = 0;
while(q){
flag = 0;
justc = C->next;
while(justc){   //重复的就不加入到该C链表中
if(q->data == justc ->data){
flag = 1;
justc = justc ->next;
}
}
if(!flag){
s = (Node *)malloc(sizeof(Node));
s ->data = q->data;
s->next = NULL;
c -> next = s;
c = c->next;
}
q = q->next;
}
while(p){
flag = 0;
justc = C->next;
while(justc){
if(c->data == p->data){
flag = 1;
}
}
if(!flag){
s = (Node *)malloc(sizeof(Node));
s -> data = p->data;
s -> next = NULL;
c-> next = s;
c = c->next;
}
p = p->next;
}
}

//计算A-B = C
void cut(Node * A,Node * B,Node * C){
Node * p = A->next;
Node * q = B->next;
Node * c = C;
Node * s;
int flag;
while(p){
flag = 0;
while(q){
if(q -> data == p -> data){
flag = 1;
}
q = q->next;
}
if(!flag){
s = (Node *)malloc(sizeof(Node));
s -> data = p -> data;
s -> next = NULL;
c -> next = s;
c = c->next;
}
p = p->next;
}
}

//找到单链表倒数第k个结点并打印，可以开始计数，当记时到链表的n个位置的时候，就让一个指针指向初始位置的结点，然后后移
struct Node * getInverseV(Node * node,int v){
Node * p = node -> next;
Node * q = NULL;
int n = 0;
while(p){
n++;
if(v == n){
q = node -> next;
}else if(n > v){
q = p -> next;
}
p = p->next;
}
return q;
}

int main(){
return 0;
}