中兴通讯2012笔试题删除双向循环链表中相同的数值

双向循环链表pHeadA ,pHeadB,

例如 pHeadA 数据为 1，2,3,4,5,7 pHeadB数据为 1,2,6,2,1,7,3

删除相同数值后为  pHeadA 4,5  pHeadB 6

#include <iostream>using namespace std;typedef struct DoubleLinkNode{int data;    struct DoubleLinkNode *pre ,*next;}node;void findx(node *pHeadA,node *pHeadB){node *pA,*pB,*pA_1,*pB_1;int x;for(pA=pHeadA->next;pA!=pHeadA;pA=pA->next){for(pB=pHeadB->next;pB!=pHeadB;pB=pB->next){if((pA->data == pB->data)){x = pA->data;cout<<"x="<<x<<endl;for(pA_1=pHeadA->next;pA_1!=pHeadA;pA_1=pA_1->next){if(pA_1->data==x){pA_1->pre->next = pA_1->next;pA_1->next->pre = pA_1->pre;pA_1 = pA_1->next; }}for(pB_1=pHeadB->next;pB_1!=pHeadB;pB_1=pB_1->next){if(pB_1->data==x){pB_1->pre->next = pB_1->next;pB_1->next->pre = pB_1->pre; pB_1 = pB_1->next;}}pA = pHeadA;}}}}void main(){node *heada,*headb,*p0,*p1,*p2,*p3,*p4,*p5,*s0,*s1,*s2,*s3,*s4,*s5;heada = (node*)malloc(sizeof(node));headb = (node*)malloc(sizeof(node));p0 = (node*)malloc(sizeof(node));p1 = (node*)malloc(sizeof(node));p2 = (node*)malloc(sizeof(node));p3 = (node*)malloc(sizeof(node));p4 = (node*)malloc(sizeof(node));p5 = (node*)malloc(sizeof(node));s0 = (node*)malloc(sizeof(node));s1 = (node*)malloc(sizeof(node));s2 = (node*)malloc(sizeof(node));s3 = (node*)malloc(sizeof(node));s4 = (node*)malloc(sizeof(node));s5 = (node*)malloc(sizeof(node)); p0->data = 1;p1->data = 2;p2->data = 5;p3->data = 4;p4->data = 3;p5->data = 7;s0->data = 1;s1->data = 5;s2->data = 1;s3->data = 7;s4->data = 2;s5->data = 3; heada->next = p0;p0->next = p1;p1->next = p2;p2->next = p3;p3->next = p4;p4->next = p5;p5->next = heada;heada->pre = p5;p5->pre = p4;p4->pre = p3;p3->pre = p2;p2->pre = p1;p1->pre = p0;p0->pre = heada;headb->next = s0;s0->next = s1;s1->next = s2;s2->next = s3;s3->next = s4;s4->next = s5;s5->next = headb;headb->pre = s5;s5->pre = s4;s4->pre = s3;s3->pre = s2;s2->pre = s1;s1->pre = s0;s0->pre = headb;    findx(heada ,headb);}

 

 

#include <iostream>using namespace std;typedef struct DoubleLinkNode{int data;    struct DoubleLinkNode *pre ,*next;}node;node *create(int n){node *head = new node();node *p =head;int x;while(n){node *s = new node();printf("请输入数据：");scanf("%d",&x);s->data = x;p->next = s;s->pre  = p;p = s;n--;}p->next = head;head->pre = p;return head;}void findx(node *pHeadA,node *pHeadB){node *pA,*pB,*pA_1,*pB_1;int x;for(pA=pHeadA->next;pA!=pHeadA;pA=pA->next){for(pB=pHeadB->next;pB!=pHeadB;pB=pB->next){if((pA->data == pB->data)){x = pA->data;cout<<"x="<<x<<endl;for(pA_1=pHeadA->next;pA_1!=pHeadA;pA_1=pA_1->next){if(pA_1->data==x) //找到链表A中的数据x，然后删除其中的数据x，{pA_1->pre->next = pA_1->next;//把节点pA_1前一个指向其pA_1的后一个pA_1->next->pre = pA_1->pre;//把pA_1的后一个节点指向的前缀指向pA_1的前一个节点//相当于把节点的前后节点进行了连接，把节点pA_1进行了孤立。pA_1 = pA_1->pre;//此时把pA_1设为它的原来的前一个节点，等到for循环时可以直接到下一个 }}for(pB_1=pHeadB->next;pB_1!=pHeadB;pB_1=pB_1->next){if(pB_1->data==x){pB_1->pre->next = pB_1->next;pB_1->next->pre = pB_1->pre; pB_1 = pB_1->pre;}}pA = pHeadA;}}}}void main(){ node *heada ,*headb;int x,y;printf("请输入个数：");scanf("%d",&x);heada = create(x);printf("请输入个数：");scanf("%d",&y);headb = create(y);         findx(heada ,headb);}