数据结构实验之链表四：有序链表的归并
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
分别输入两个有序的整数序列（分别包含M和N个数据），建立两个有序的单链表，将这两个有序单链表合并成为一个大的有序单链表，并依次输出合并后的单链表数据。
Input
第一行输入M与N的值；
第二行依次输入M个有序的整数；
第三行依次输入N个有序的整数。
Output
输出合并后的单链表所包含的M+N个有序的整数。
Sample Input
6 5
1 23 26 45 66 99
14 21 28 50 100
Sample Output
1 14 21 23 26 28 45 50 66 99 100
Hint
不得使用数组！
Source
#include <stdio.h> #include <stdlib.h> struct node { int data; struct node *next; }; int main() { struct node *head, *head1, *head2, *tail1, *tail2, *tail, *p; int m, n, i; scanf("%d%d", &m, &n); head1 = (struct node *)malloc(sizeof(struct node)); head1->next = NULL; tail = head1; for(i = 1; i <= m; i++) { p = (struct node *)malloc(sizeof(struct node)); scanf("%d", &p->data); p->next = NULL; tail->next = p; tail = p; } head2 = (struct node *)malloc(sizeof(struct node)); head2->next = NULL; tail = head2; for(i = 1; i <= n; i++) { p = (struct node *)malloc(sizeof(struct node)); scanf("%d", &p->data); p->next = NULL; tail->next = p; tail = p; } head = (struct node *)malloc(sizeof(struct node)); head->next = NULL; tail = head; tail1 = head1->next; tail2 = head2->next; for(i = 1; i <= m + n; i++) { if(tail1 == NULL) { tail->next = tail2; tail = tail2; tail2 = tail2->next; } else if(tail2 == NULL) { tail->next = tail1; tail = tail1; tail1 = tail1->next; } else { if(tail1->data < tail2->data) { tail->next = tail1; tail = tail1; tail1 = tail1->next; } else { tail->next = tail2; tail = tail2; tail2 = tail2->next; } } } tail->next = NULL; for(tail = head->next; tail->next != NULL; tail = tail->next) { printf("%d ", tail->data); } printf("%d\n", tail->data); return 0; }
