C语言设计车牌号基数链表排序,并支持关键字折半查找。初始界面按1是人工录入,并跳转录入界面,录入完成后,支持查找。初始界面按2为随机数自动生成,生成完成后,支持查找
时间: 2024-03-15 21:46:13 浏览: 16
好的,这是一个比较复杂的问题,我会尽力回答。
首先,我们需要定义一个车牌号基数链表的数据结构,它包含以下字段:
```c
typedef struct LicensePlate {
char code[10]; // 车牌号码,字符串类型
struct LicensePlate *next; // 指向下一个节点的指针
} LicensePlate;
```
接下来,我们需要实现一个函数用于在车牌号基数链表中插入一个节点,这个函数的实现如下:
```c
void insert(LicensePlate **head, char *code) {
LicensePlate *new_node = (LicensePlate *) malloc(sizeof(LicensePlate));
strcpy(new_node->code, code);
new_node->next = NULL;
if (*head == NULL) {
*head = new_node;
} else {
LicensePlate *current = *head;
while (current->next != NULL) {
current = current->next;
}
current->next = new_node;
}
}
```
我们还需要实现一个函数,用于按照车牌号基数对链表进行排序,这个函数的实现如下:
```c
void radix_sort(LicensePlate **head) {
int i, j;
LicensePlate *list[10], *current;
for (i = 0; i < 10; i++) {
list[i] = NULL;
}
int max_len = 0;
current = *head;
while (current != NULL) {
int len = strlen(current->code);
if (len > max_len) {
max_len = len;
}
current = current->next;
}
for (i = 0; i < max_len; i++) {
current = *head;
while (current != NULL) {
int index = current->code[max_len - i - 1] - '0';
if (list[index] == NULL) {
list[index] = current;
*head = current->next;
current->next = NULL;
} else {
LicensePlate *temp = list[index];
while (temp->next != NULL) {
temp = temp->next;
}
temp->next = current;
current = current->next;
temp->next->next = NULL;
}
}
for (j = 0; j < 10; j++) {
if (list[j] != NULL) {
current = list[j];
while (current->next != NULL) {
current = current->next;
}
current->next = *head;
*head = list[j];
list[j] = NULL;
}
}
}
}
```
现在我们已经实现了车牌号基数链表的排序,接下来是关键字折半查找。我们可以实现一个函数,用于在排序后的链表中查找指定的车牌号码,这个函数的实现如下:
```c
LicensePlate *binary_search(LicensePlate *head, char *code) {
int low = 0, high = count(head) - 1, mid;
while (low <= high) {
mid = (low + high) / 2;
LicensePlate *current = head;
int i;
for (i = 0; i < mid; i++) {
current = current->next;
}
if (strcmp(current->code, code) == 0) {
return current;
} else if (strcmp(current->code, code) < 0) {
low = mid + 1;
} else {
high = mid - 1;
}
}
return NULL;
}
```
其中,count() 函数用于计算链表中节点的个数,其实现如下:
```c
int count(LicensePlate *head) {
int count = 0;
LicensePlate *current = head;
while (current != NULL) {
count++;
current = current->next;
}
return count;
}
```
最后,我们需要实现两个界面,一个是人工录入界面,一个是随机数自动生成界面。这个可以使用 switch 语句实现,具体实现如下:
```c
int main() {
LicensePlate *head = NULL;
int choice;
while (1) {
printf("请输入选项:\n");
printf("1. 人工录入\n");
printf("2. 随机数自动生成\n");
printf("3. 退出\n");
scanf("%d", &choice);
switch (choice) {
case 1:
// 人工录入界面
int n;
printf("请输入车牌号码的个数:\n");
scanf("%d", &n);
while (n--) {
char code[10];
printf("请输入车牌号码:\n");
scanf("%s", code);
insert(&head, code);
}
break;
case 2:
// 随机数自动生成界面
int m;
printf("请输入车牌号码的个数:\n");
scanf("%d", &m);
while (m--) {
char code[10];
sprintf(code, "%06d", rand() % 1000000);
insert(&head, code);
}
break;
case 3:
// 退出
exit(0);
default:
printf("无效的选项!\n");
break;
}
radix_sort(&head);
char code[10];
printf("请输入要查找的车牌号码:\n");
scanf("%s", code);
LicensePlate *result = binary_search(head, code);
if (result == NULL) {
printf("未找到车牌号码为 %s 的车辆!\n", code);
} else {
printf("找到车牌号码为 %s 的车辆!\n", code);
}
}
return 0;
}
```
以上就是 C 语言设计车牌号基数链表排序并支持关键字折半查找的实现。