單向鏈表逆序問題

問題：將一個單向單向鏈表逆序倒轉，要求O(n)的時間複雜度和O(1)的空間複雜度

該問題有兩種解法，一種非遞歸，一種是遞歸，代碼如下：

#include <stdio.h>
#include "Link.h"

typedef struct linknode
{
	int value;
	struct linknode* next;
} LinkNode;

//遞歸
LinkNode* ReverseLink(LinkNode* link)
{
	if (link->next == NULL)
	{
		return link;
	}
	LinkNode* nextNode = link->next;
	LinkNode* reverseLink = ReverseLink(link->next);
	nextNode->next = link;
	link->next = NULL;
	return reverseLink;
}

//非遞歸
LinkNode* ReverseLink2(LinkNode* link)
{
	LinkNode* pre = NULL;
	LinkNode* curNode = link;
	LinkNode* tempNode = NULL;
	while (curNode)
	{
		tempNode = curNode->next;
		curNode->next = pre;
		pre = curNode;
		curNode = tempNode;
	}
	LinkNode* reverseLink = pre;
	return reverseLink;
}

int main(int argc, char** argv)
{
	LinkNode* head = NULL;
	LinkNode* tail = NULL;
	for (int i = 101; i < 200; i += 5)
	{
		LinkNode* p = new LinkNode;
		p->value = i;
		p->next = NULL;
		if (!head)
		{
			head = p;
			tail = head;
		}
		else
		{
			tail->next = p;
			tail = p;
		}
	}
	LinkNode* publicLink = head;
	printf("Before Reverse: ");
	LinkNode* temp = publicLink;
	while (temp)
	{
		printf("%d ", temp->value);
		temp = temp->next;
	}
	printf("\n");

	printf("After Reverse: ");
	//LinkNode* reverseLink = ReverseLink(publicLink);
	LinkNode* reverseLink = ReverseLink2(publicLink);
	temp = reverseLink;
	while (temp)
	{
		printf("%d ", temp->value);
		temp = temp->next;
	}
	getchar();
	return 0;
}