合并两个链表的问题

来源：互联网发布：淘宝网外衣编辑：程序博客网时间：2024/04/30 02:12
#include <stdio.h>#include <stdlib.h>#include <time.h>#include <conio.h>#define N 10typedef struct Node{int data;struct Node *next;}Node, *pNode;/*顺序插入法*/void create_list_sequence(pNode *h){pNode p, q, r=NULL;p = q = *h = (pNode)calloc(1,sizeof(Node));p->next = NULL;int count = 0;while (count != N){++count;if (count == 1){p->data = rand()%100;//printf("%d ", p->data);}else{r = (pNode)calloc(1,sizeof(Node));r->data = rand() % 100;//printf("%d ", r->data);p = q = *h;while (p ->next != NULL && p->data < r->data ){q = p;p = p->next;}if (p->data >= r->data){if (p == q){r->next = *h;*h = r;}else {r->next = p;q->next = r;}}else{p->next = r;r->next = NULL;}}}printf("\n");}/*方法一：在这个合并俩链表的程序中，除了刚开始，pa1,pa2始终是一前一后，pb1和pb2始终是b中当前所剩节点的头节点*/void mergeTwoList(pNode *h1, pNode *h2){pNode pa1, pa2, pb1, pb2;pa1 = pa2 = *h1;pb1 = pb2 = *h2;while (pa1 != NULL&&pb1 != NULL){/*当两个链表其中任何一个遍历结束，就退出循环*/while (pb1->data > pa1->data && pa1->next != NULL){/*找到a链表中不小于b链表中元素的元素，如果一直没有，则到a中的最后一个元素为止*/pa2 = pa1;pa1 = pa1->next;}if (pa1->data >= pb1->data){/*找到了a链表中不小于b链表中元素的元素*/if (pa1 == pa2){/*如果是a聊表的第一个元素，就把a链表的头指针指向b链表的第一个节点*/*h1 = pb1;}else{pa2->next = pb1;/*如果不是a链表中的第一个元素，则让a链表当前节点的上一个节点指向b当前节点*/}pb1 = pb1->next;/*将b1向后移动一个位子，为pb2指向pa1，从而防止会和b表中下一个节点失去联系*/pb2->next = pa1;pa2 = pb2;pb2 = pb1;}else{pa2 = pa1;/*如果a当前元素比b小，且是a中最后一个元素，就往后移动，其实就会执行一次，只是是为了解决死循环问题*/pa1 = pa1->next;}}if (pa1 == NULL && pb1 != NULL){pa2->next = pb1;}}void mergeTwoList_test(){pNode Link1, Link2;create_list_sequence(&Link1);create_list_sequence(&Link2);printf("Link1:\n");printList(Link1);printf("Link2:\n");printList(Link2);mergeTwoList(&Link1, &Link2);printf("After merge:\n");printList(Link1);}/*方法二：*/void mergeTwoList_solution2(pNode *h1,pNode *h2){pNode pa1, pa2, pb1, pb2;pa1 = pa2 = *h1;pb1 = pb2 = *h2;while (pa1 != NULL && pb1 != NULL){if (pa1->data > pb1->data){if (pa1 == pa2){*h1 = pb1;pa2 = pb1;}else{pa2->next = pb1;pa2 = pa2->next;pb1 = pb1->next;}}else if (pa1->data < pb1->data){if (pa1 != pa2){pa2->next = pa1;pa2 = pa2->next;}pa1 = pa1->next;}else if (pa1->data == pb1->data){if (pa1 != pa2){pa2->next = pa1;pa2 = pa2->next;}pa1 = pa1->next;pa2->next = pb1;pa2 = pa2->next;pb1 = pb1->next;}}if (pa1 == NULL){pa2->next = pb1;}if (pb1 == NULL){pa2->next = pa1;}}int main(){mergeTwoList_test();return 0;}
0 0