8---------单链表算法的简单应用和巩固

单链表算法主要注意处理好指针的指向问题
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

#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 = Cvoid 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;}