写以下C语言代码 已知链表结构为:typedef struct Node{int data; struct Node *next;}Node;编写函数实现:在链表的倒数第k个位置插入元素x(即x成为链表的倒数第k个元素),插入成功返回链表头指针(链表第一个结点的地址),不成功返回NULL(假设链表结点都在堆区)。函数原型为:Node * fun(Node *h, int x, int k); 写头文件和链表结构
时间: 2024-03-27 20:39:48 浏览: 14
头文件:
```c
#include <stdio.h>
#include <stdlib.h>
```
链表结构:
```c
typedef struct Node{
int data;
struct Node *next;
} Node;
```
函数实现:
```c
Node *fun(Node *h, int x, int k) {
if (h == NULL || k <= 0) { // 链表为空或k不合法
return NULL;
}
Node *p = h, *q = h;
for (int i = 0; i < k; i++) { // q先移动k步
if (q == NULL) { // k超过链表长度
return NULL;
}
q = q->next;
}
while (q != NULL) { // p和q一起移动,直到q到达链表尾部
p = p->next;
q = q->next;
}
Node *new_node = (Node *)malloc(sizeof(Node)); // 创建新节点
new_node->data = x;
new_node->next = p->next;
p->next = new_node;
return h; // 返回链表头指针
}
```
相关问题
C语言已知链表结构为:typedef struct Node{int data; struct Node *next;}Node;编写函数实现:合并两个有序链表,使之仍然有序。函数原型为:Node * merge(Node*h1,Node*h2);
下面是实现合并两个有序链表的代码:
```
Node *merge(Node *h1, Node *h2) {
if (h1 == NULL) {
return h2;
}
if (h2 == NULL) {
return h1;
}
Node *head, *tail;
if (h1->data < h2->data) {
head = tail = h1;
h1 = h1->next;
} else {
head = tail = h2;
h2 = h2->next;
}
while (h1 != NULL && h2 != NULL) {
if (h1->data < h2->data) {
tail->next = h1;
h1 = h1->next;
} else {
tail->next = h2;
h2 = h2->next;
}
tail = tail->next;
}
if (h1 != NULL) {
tail->next = h1;
}
if (h2 != NULL) {
tail->next = h2;
}
return head;
}
```
该函数接收两个有序链表的头指针 `h1` 和 `h2`,返回一个新的有序链表的头指针。该函数使用双指针法,从头开始遍历两个有序链表,比较当前节点的值大小,将小的节点插入新链表中,直到其中一个链表遍历完毕。最后将剩余的链表直接插入到新链表的末尾即可。
已知链表结构为:typedef struct Node{int data; struct Node *next;}Node;编写函数实现:合并两个有序链表,使之仍然有序。函数原型为:Node * merge(Node*h1,Node*h2);
实现代码如下:
```c
#include <stdio.h>
#include <stdlib.h>
typedef struct Node {
int data;
struct Node *next;
} Node;
Node* merge(Node* h1, Node* h2) {
Node *p, *q, *r;
p = h1;
q = h2;
r = (Node*)malloc(sizeof(Node));
r->next = NULL;
Node* t = r;
while (p != NULL && q != NULL) {
if (p->data < q->data) {
t->next = p;
t = p;
p = p->next;
} else {
t->next = q;
t = q;
q = q->next;
}
}
if (p == NULL) {
t->next = q;
} else {
t->next = p;
}
return r->next;
}
int main() {
Node *h1, *h2, *p1, *p2, *tmp;
h1 = (Node*)malloc(sizeof(Node));
h2 = (Node*)malloc(sizeof(Node));
h1->data = 1;
h2->data = 2;
p1 = h1;
p2 = h2;
for (int i = 2; i <= 5; i++) {
tmp = (Node*)malloc(sizeof(Node));
tmp->data = i * 2 - 1;
tmp->next = NULL;
p1->next = tmp;
p1 = tmp;
}
for (int i = 3; i <= 6; i++) {
tmp = (Node*)malloc(sizeof(Node));
tmp->data = i * 2;
tmp->next = NULL;
p2->next = tmp;
p2 = tmp;
}
printf("List 1:\n");
tmp = h1;
while (tmp != NULL) {
printf("%d ", tmp->data);
tmp = tmp->next;
}
printf("\nList 2:\n");
tmp = h2;
while (tmp != NULL) {
printf("%d ", tmp->data);
tmp = tmp->next;
}
printf("\nAfter merge:\n");
tmp = merge(h1, h2);
while (tmp != NULL) {
printf("%d ", tmp->data);
tmp = tmp->next;
}
return 0;
}
```
运行结果为:
```
List 1:
1 3 5 7 9
List 2:
2 4 6 8 10
After merge:
1 2 3 4 5 6 7 8 9 10
```