数据结构:单向链表系列8--反转链表
转载
业务需求:给定一个指向头指针的链表,反转链表。实现过程:更改相邻节点之间的链域。
例:
输入:
1->2->3->4->NULL
输出:
4->3->2->1->NULL
输入:
1->2->3->4->5->NULL
输出:
5->4->3->2->1->NULL
输入: NULL
输出: NULL
输入: 1->NULL
输出: 1->NULL
迭代法:
空间复杂度:O(1),时间复杂度:O(n)
1、初始化3个指针
pre = NULL, curr = *head, next = NULL
2、迭代列表,执行以下循环
// Before changing next of current,
// store next node
next = curr->next
// Now change next of current
// This is where actual reversing happens
curr->next = prev
// Move prev and curr one step forward
prev = curr
curr = next
以下是算法实现过程:
// Iterative C program to reverse a linked list
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
struct Node
{
int data;
struct Node* next;
};
/* function to reverse the linked list */
void reverse(struct Node** head_ref)
{
struct Node* prev = NULL;
struct Node* current = *head_ref;
struct Node* next = NULL;
while(current != NULL)
{
//Store next
next = current->next;
//Reverse current node's pointer
current->next = prev;
//Move pointers one position ahead.
prev = current;
current = next;
}
*head_ref = prev;
}
/* function to push a node */
void push(struct Node** head_ref, int new_data)
{
struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
/* function to print linked list */
void printList(struct Node* head)
{
struct Node* temp = head;
while(temp != NULL)
{
printf("%d ", temp->data);
temp = temp->next;
}
printf("\n");
}
/* driver program to test above function */
int main() {
/* Start with the empty list */
struct Node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
printf("Given linked list\n");
printList(head);
reverse(&head);
printf("Reversed linked list\n");
printList(head);
return 0;
}