数据结构c语言 单链表的就地逆置

单链表的就地逆置是指不引入新的节点，只调整原有节点的指针，将单链表逆序排列。

实现单链表的就地逆置可以使用三个指针：prev、curr和next。开始时，将prev指向NULL，curr指向链表的首节点，next指向curr的下一个节点。

接下来，通过循环遍历单链表，每次将curr的指针指向prev，然后将prev、curr和next指针都向后移动一个节点。直到curr指向NULL，表示已经遍历到链表的末尾。

最后，将链表的头节点指向prev，完成就地逆置。

以下是C语言实现的代码示例：

#include <stdio.h>

struct Node {
int data;
struct Node* next;
};

// 单链表就地逆置函数
void reverseList(struct Node** head){
struct Node* prev = NULL;
struct Node* curr = *head;
struct Node* next = NULL;

while(curr != NULL){
next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}

*head = prev;
}

// 打印单链表函数
void printList(struct Node* head){
struct Node* temp = head;
while(temp != NULL){
printf("%d ", temp->data);
temp = temp->next;
}

printf("\n");
}

int main(){
// 创建一个示例单链表: 1 -> 2 -> 3 -> 4 -> 5
struct Node* head = NULL;
struct Node* second = NULL;
struct Node* third = NULL;
struct Node* fourth = NULL;
struct Node* fifth = NULL;

head = (struct Node*)malloc(sizeof(struct Node));
second = (struct Node*)malloc(sizeof(struct Node));
third = (struct Node*)malloc(sizeof(struct Node));
fourth = (struct Node*)malloc(sizeof(struct Node));
fifth = (struct Node*)malloc(sizeof(struct Node));

head->data = 1;
head->next = second;
second->data = 2;
second->next = third;
third->data = 3;
third->next = fourth;
fourth->data = 4;
fourth->next = fifth;
fifth->data = 5;
fifth->next = NULL;

printf("原始链表: ");
printList(head);

reverseList(&head);

printf("就地逆置后的链表: ");
printList(head);

return 0;
}

运行此程序，输出结果为：

原始链表: 1 2 3 4 5
就地逆置后的链表: 5 4 3 2 1