02-线性结构1 两个有序链表序列的合并 (15分)

本题要求实现一个函数，将两个链表表示的递增整数序列合并为一个非递减的整数序列。

函数接口定义：

List Merge( List L1, List L2 );

其中

List

结构定义如下：

typedef struct Node *PtrToNode;
struct Node {
ElementType Data; /* 存储结点数据 */
PtrToNode   Next; /* 指向下一个结点的指针 */
};
typedef PtrToNode List; /* 定义单链表类型 */

L1

和

L2

是给定的带头结点的单链表，其结点存储的数据是递增有序的；函数

Merge

要将

L1

和

L2

合并为一个非递减的整数序列。应直接使用原序列中的结点，返回归并后的链表头指针。

完整的程序如下：但是题目只是让提交子函数，要注意

#include <stdio.h>
#include <stdlib.h>

typedef int ElementType;
typedef struct Node *PtrToNo
9642
de;
struct Node {
ElementType Data;
PtrToNode Next;
};
typedef PtrToNode List;

List Read();
List Insert(List L, ElementType X); // 在链尾插入
void Print(List L);

List Merge(List L1, List L2);

int main()
{
List L1, L2, L;
L1 = Read();
L2 = Read();
L = Merge(L1, L2);
Print(L);
Print(L1);
Print(L2);

return 0;
}

List Read()
{
int num = 0;
List Head;
PtrToNode last;
Head = (PtrToNode)malloc(sizeof(struct Node));
Head->Next = NULL;
last = Head;
scanf("%d", &num);
if (num == 0)
return NULL;

while (num--)
{
List tmp = (PtrToNode)malloc(sizeof(struct Node));
scanf("%d", &tmp->Data);
last->Next = tmp;
tmp->Next = NULL;
last = tmp;
}
return Head;
}

List Merge(List L1, List L2)
{
PtrToNode Last;
PtrToNode Head = (PtrToNode)malloc(sizeof(struct Node));
List L1_tmp;
List L2_tmp;
Head->Next = NULL;
Last = Head;

L1_tmp = L1->Next;
L2_tmp = L2->Next;

for (; L1_tmp != NULL&&L2_tmp != NULL;)
{
if (L1_tmp->Data < L2_tmp->Data)
{
Last->Next = L1_tmp;
Last = L1_tmp;
L1_tmp = L1_tmp->Next;
}
else if (L1_tmp->Data > L2_tmp->Data)
{
Last->Next = L2_tmp;
Last = L2_tmp;
L2_tmp = L2_tmp->Next;
}
else
{
Last->Next = L1_tmp;
Last = L1_tmp;
L1_tmp = L1_tmp->Next;
Last->Next = L2_tmp;
Last = L2_tmp;
L2_tmp = L2_tmp->Next;
}
}
if (L1_tmp == NULL)
{
Last->Next = L2_tmp;
}
else if (L2_tmp == NULL)
{
Last->Next = L1_tmp;
}
L1->Next = NULL;
L2->Next = NULL;
return Head;
}
void Print(List L)
{
if (L->Next == NULL){
printf("NULL\n");
return;
}
L = L->Next;
for (; L != NULL;){
printf("%d ", L->Data);
L = L->Next;
}
putchar('\n');
}