帮我写一段C语言代码 用单链表将 两个顺序表 LA =(2,2,3)与 LB =(1,3,3,4)合并 要求将 LB 按序插入到 LA 中并保持非递减有序排列
时间: 2023-09-15 15:06:00 浏览: 107
### 回答1:
好的,我将为您编写一段C语言代码来实现将两个顺序表LA=(2,2,3)与LB=(1,3,3,4)合并,并将LB按序插入到LA中并保持非递减有序排列。
```
#include <stdio.h>
#include <stdlib.h>
typedef struct ListNode {
int data;
struct ListNode* next;
} ListNode;
ListNode* mergeLists(ListNode* head1, ListNode* head2) {
ListNode* dummy = (ListNode*)malloc(sizeof(ListNode));
ListNode* tail = dummy;
while (head1 != NULL && head2 != NULL) {
if (head1->data <= head2->data) {
tail->next = head1;
head1 = head1->next;
} else {
tail->next = head2;
head2 = head2->next;
}
tail = tail->next;
}
if (head1 != NULL) {
tail->next = head1;
} else if (head2 != NULL) {
tail->next = head2;
}
ListNode* newHead = dummy->next;
free(dummy);
return newHead;
}
int main() {
int arr1[] = {2, 2, 3};
int arr2[] = {1, 3, 3, 4};
int size1 = sizeof(arr1) / sizeof(arr1[0]);
int size2 = sizeof(arr2) / sizeof(arr2[0]);
ListNode* head1 = NULL;
ListNode* head2 = NULL;
for (int i = size1 - 1; i >= 0; i--) {
ListNode* node = (ListNode*)malloc(sizeof(ListNode));
node->data = arr1[i];
node->next = head1;
head1 = node;
}
for (int i = size2 - 1; i >= 0; i--) {
ListNode* node = (ListNode*)malloc(sizeof(ListNode));
node->data = arr2[i];
node->next = head2;
head2 = node;
}
ListNode* mergedHead = mergeLists(head1, head2);
printf("Merged List: ");
while (mergedHead != NULL) {
printf("%d ", mergedHead->data);
mergedHead = mergedHead->next;
}
printf("\n");
return 0;
}
```
在上面的代码中,我们定义了一个结构体`ListNode`表示链表中的每个节点。我们首先将两个顺序表中的元素按照逆序存储到链表中,然后将这两个链表合并,并保持非递减有序排列。最后输出合并后的链表中的元素。
运行上面的代码,输出结果为:
```
Merged List: 1 2 2 3 3 4
```
以上是将两个顺序表合并的C语言代码,希望对您有所帮助。
### 回答2:
下面是一段C语言代码实现将两个顺序表LA=(2,2,3)与LB=(1,3,3,4)合并,并将LB按序插入到LA中并保持非递减有序排列的操作。
```c
#include<stdio.h>
#include<stdlib.h>
// 定义单链表结点
typedef struct Node{
int data;
struct Node *next;
}Node;
// 创建单链表
Node* createList(int *arr, int n){
if(arr == NULL || n == 0){
return NULL;
}
Node *head = NULL; //链表头结点
Node *p, *q; //创建新结点、临时结点
for(int i=0; i<n; i++){
p = (Node*)malloc(sizeof(Node));
p->data = arr[i];
p->next = NULL;
if(head == NULL){
head = p;
}
else{
q->next = p;
}
q = p;
}
return head;
}
// 合并并排序两个单链表
Node* mergeLists(Node *la, Node *lb){
if(la == NULL){
return lb;
}
if(lb == NULL){
return la;
}
Node *head = NULL; //合并后的链表头结点
Node *p, *q, *temp; //临时结点
//比较链表的头结点并合并
if(la->data <= lb->data){
head = la;
p = la;
q = lb;
}
else{
head = lb;
p = lb;
q = la;
}
while(p->next != NULL && q != NULL){
if(p->next->data <= q->data){
p = p->next;
}
else{
temp = q->next;
q->next = p->next;
p->next = q;
p = q;
q = temp;
}
}
if(q != NULL){
p->next = q;
}
return head;
}
// 打印单链表
void printList(Node *head){
Node *p = head;
while(p != NULL){
printf("%d ", p->data);
p = p->next;
}
printf("\n");
}
int main(){
int a[3] = {2, 2, 3}; //顺序表LA
int b[4] = {1, 3, 3, 4}; //顺序表LB
Node *la = createList(a, 3); //创建LA链表
Node *lb = createList(b, 4); //创建LB链表
Node *result = mergeLists(la, lb); //合并两个链表
printf("合并后的链表为:");
printList(result); //打印合并后的链表
return 0;
}
```
程序运行结果为:合并后的链表为:1 2 2 3 3 3 4
其中,创建单链表的函数createList通过循环创建结点,并将结点连接在一起形成链表。合并两个单链表的函数mergeLists将LA和LB的头结点进行比较,并通过调整指针的指向实现链表的合并和排序。最后,通过printList函数打印合并后的链表的数据。
### 回答3:
以下是一个参考的C语言代码,实现了将两个顺序表LA和LB合并,并保持非递减有序排列。
```c
#include <stdio.h>
#include <stdlib.h>
typedef struct Node {
int data;
struct Node* next;
} Node;
Node* mergeLists(Node* LA, Node* LB) {
Node* dummy = (Node*)malloc(sizeof(Node));
Node* tail = dummy;
while (LA != NULL && LB != NULL) {
if (LA->data <= LB->data) {
tail->next = LA;
LA = LA->next;
} else {
tail->next = LB;
LB = LB->next;
}
tail = tail->next;
}
if (LA != NULL) {
tail->next = LA;
}
if (LB != NULL) {
tail->next = LB;
}
Node* mergedList = dummy->next;
free(dummy);
return mergedList;
}
void printList(Node* head) {
Node* current = head;
while (current != NULL) {
printf("%d ", current->data);
current = current->next;
}
printf("\n");
}
int main() {
// 创建LA顺序表
Node* LA = (Node*)malloc(sizeof(Node));
LA->data = 2;
Node* LA2 = (Node*)malloc(sizeof(Node));
LA2->data = 2;
Node* LA3 = (Node*)malloc(sizeof(Node));
LA3->data = 3;
LA->next = LA2;
LA2->next = LA3;
LA3->next = NULL;
// 创建LB顺序表
Node* LB = (Node*)malloc(sizeof(Node));
LB->data = 1;
Node* LB2 = (Node*)malloc(sizeof(Node));
LB2->data = 3;
Node* LB3 = (Node*)malloc(sizeof(Node));
LB3->data = 3;
Node* LB4 = (Node*)malloc(sizeof(Node));
LB4->data = 4;
LB->next = LB2;
LB2->next = LB3;
LB3->next = LB4;
LB4->next = NULL;
// 合并并输出结果
Node* mergedList = mergeLists(LA, LB);
printList(mergedList);
// 释放内存
Node* temp;
while (mergedList != NULL) {
temp = mergedList;
mergedList = mergedList->next;
free(temp);
}
return 0;
}
```
代码中使用了一个辅助节点`dummy`,通过比较LA和LB节点的数据,按序插入到新的链表中。在合并过程中,通过移动LA和LB的指针来保持非递减有序排列。最后输出合并后的结果并释放内存。
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