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 *PtrToNode; 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'); }