鏈表翻轉

要求用儘可能快的方式實現鏈表的翻轉操作。

這個我們需要用兩個指針，一個指向當前的節點，一個指向當前節點的前一個節點，每一次使當前節點的指向前一個節點來實現兩個節點之間的翻轉，然後順次再移動實現循環。

代碼也很簡單，如下所示：

#include <iostream>
using namespace std;

struct Node
{
	int key;
	Node* next;
};
Node* createList(int arr[],int nLength);
Node* reverseList(Node* head);
void printList(Node* head);
void clearList(Node* head);

void main()
{
	int arr[] = {1,3,5,7,9};
	int nLength = sizeof(arr)/sizeof(arr[0]);
	Node* head = createList(arr,nLength);
	printList(head);
	head = reverseList(head);
	printList(head);
	clearList(head);
}

Node* createList(int arr[],int nLength)
{
	Node* head = new Node;
	head->key = arr[0];
	head->next = NULL;
	Node *p = head;
	for(int i=1;i<nLength;i++)
	{
		Node* ptr = new Node;
		ptr->key = arr[i];
		ptr->next = NULL;
		p->next = ptr;
		p = p->next;
	}
	return head;
}

Node* reverseList(Node* head)
{
	Node* preNode = NULL;
	Node* pNode = head;
	while( pNode != NULL )
	{
		Node* pNext = pNode->next;
		pNode->next = preNode;
		preNode = pNode;
		pNode = pNext;

	}
	return preNode;
}

void printList(Node* head)
{
	Node* p = head;
	while( p!= NULL )
	{
		cout<<p->key<<endl;
		p=p->next;
	}
}

void clearList(Node* head)
{
	Node* p = head;
	Node* ptr;
	while( p!= NULL )
	{
		ptr = p->next;
		delete p;
		p = ptr;
	}
}