用C或C++编程不使用函数方法实现顺序表,要求顺序表的定义至少包含(1)创建顺序表(2)顺序表的按位查找(3)顺序表的按位插入和按位删除(4)按位获取某个元素等功能;然后实现两个顺序表的并操作(不允许重复元素)和两个有序顺序表的合并操作(允许重复元素)。
时间: 2024-09-23 18:08:11 浏览: 34
使用C++实现顺序链表
在C或C++中,不直接使用函数来实现顺序表,我们可以手动定义结构体或类来表示列表,并提供相应的成员函数来操作。以下是一个简单的顺序表定义示例:
```c++
#include <stdbool.h>
// 定义顺序表节点
struct ListNode {
int data;
struct ListNode* next;
};
// 创建空顺序表
ListNode* createList() {
return (ListNode*)malloc(sizeof(ListNode));
}
// 按位查找
bool search(int target, ListNode* head) {
ListNode* current = head;
while (current != NULL) {
if (current->data == target)
return true;
current = current->next;
}
return false;
}
// 按位插入
void insert(int value, ListNode** head) {
ListNode* newNode = createList();
newNode->data = value;
newNode->next = *head;
*head = newNode;
}
// 按位删除
void remove(int value, ListNode** head) {
ListNode* prev = NULL;
ListNode* current = *head;
while (current != NULL && current->data != value) {
prev = current;
current = current->next;
}
if (current != NULL) {
if (prev == NULL)
*head = current->next;
else
prev->next = current->next;
free(current);
}
}
// 获取元素
int get(int index, ListNode* head) {
ListNode* current = head;
for (int i = 0; i < index && current != NULL; i++, current = current->next)
;
return current ? current->data : -1; // 如果超过范围,返回-1
}
// 并操作(不允许重复元素)
ListNode* mergeDisjointLists(ListNode* list1, ListNode* list2) {
if (!list1 || !list2) return list1 ? list1 : list2;
ListNode* merged = createList();
ListNode* temp1 = list1;
ListNode* temp2 = list2;
while (temp1 && temp2) {
if (temp1->data < temp2->data) {
merged->next = temp1;
temp1 = temp1->next;
} else if (temp1->data > temp2->data) {
merged->next = temp2;
temp2 = temp2->next;
} else { // 避免重复元素
temp1 = temp1->next;
temp2 = temp2->next;
}
merged = merged->next;
}
merged->next = temp1 ? temp1 : temp2;
return merged;
}
// 合并有序列表(允许重复元素)
ListNode* mergeOrderedLists(ListNode* list1, ListNode* list2) {
if (!list1) return list2;
if (!list2) return list1;
ListNode* merged = createList();
merged->next = list1;
ListNode* current = list1;
while (list2 != NULL) {
if (current->data >= list2->data) {
merged->next = list2;
list2 = list2->next;
} else {
merged->next = current;
current = current->next;
}
merged = merged->next;
}
if (current != NULL)
merged->next = current;
return merged;
}
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