数据结构实验之链表四:有序链表的归并
Time Limit: 1000MS Memory limit: 65536K
题目描述
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。
输入
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。
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。
输出
输出合并后的单链表所包含的M+N个有序的整数。
示例输入
6 5 1 23 26 45 66 99 14 21 28 50 100
示例输出
1 14 21 23 26 28 45 50 66 99 100
提示
不得使用数组!
#include<stdio.h> #include<stdlib.h> struct node { int data; struct node *next; }; struct node *merge(struct node *head1, struct node *head2) { struct node *tail, *p1, *p2; p1 = head1->next; p2 = head2->next; tail = head1; free(head2); while(p1 && p2) { if(p1->data < p2->data) { tail->next = p1; tail = p1; p1 = p1->next; } else { tail->next = p2; tail = p2; p2 = p2->next; } } if(p1) tail->next = p1; else tail->next = p2; return (head1); } int main() { int n, m, i; struct node *head1, *head2, *tail1, *tail2, *p1, *p2, *turn; head1 = (struct node *)malloc(sizeof(struct node)); head2 = (struct node *)malloc(sizeof(struct node)); tail1 = head1; tail2 = head2; head1->next = NULL; head2->next = NULL; scanf("%d %d", &n, &m); for(i=0; i<n; i++) { p1 = (struct node *)malloc(sizeof(struct node)); scanf("%d", &p1->data); tail1->next = p1; p1->next = NULL; tail1 = p1; } for(i=0; i<m; i++) { p2 = (struct node *)malloc(sizeof(struct node)); scanf("%d", &p2->data); tail2->next = p2; tail2 = p2; p2->next = NULL; } turn = merge(head1, head2); turn = turn->next; for(i=0; i<n+m; i++) { printf("%d", turn->data); if(i==n+m-1) printf(" "); else printf(" "); turn = turn->next; } return 0; }