鏈表逆置

單鏈表逆置問題:

1頭插法

2遞歸

具體過程大致圖如下:

代碼實現:

#include <iostream>
using namespace std;

typedef int elemtype;
typedef struct ListNode
{
	elemtype data;
	ListNode *next;
}ListNode,*List;
ListNode *Buynode()
{
	ListNode* p = (ListNode *)malloc(sizeof(ListNode));
	if (NULL == p)
		exit (-1);
	memset(p,0,sizeof(ListNode));
	p->next = NULL;
	return p;
}
ListNode *init_list()
{
	ListNode *p = Buynode();
	return p;
}

void push_front(ListNode *head,int value)
{
	if (head == NULL)
	{
		return;
	}
	ListNode*p = Buynode();
	p->data = value;
	p->next = head->next;
	head->next = p;

}

void print(ListNode *head)
{
	if (head == NULL)
		return ;
	ListNode *pcur = head->next;
	while (pcur != NULL)
	{
		cout<<pcur->data<<" ";
		pcur = pcur->next;
	}
	cout<<endl;
}
////////////鏈表逆置--頭插法////////////////////
void rerverse(ListNode *head)
{
	if (head == NULL)
		return ;
	ListNode *p = head->next;
	ListNode *s = NULL;
	head->next = NULL;
	while (p!= NULL)//頭插s,改變頭結點的指向
	{
		s = p;
		p = p->next;
		s->next = head->next;
		head->next = s;
	}
}
////////////////////////鏈表逆置 (遞歸)//////////////////////////
ListNode *phead = NULL;//用來記錄逆置之後新的第一個結點
void reverse2(ListNode *s,ListNode *p)
{
	if (p != NULL)
	{
		reverse2(p,p->next);
		p->next = s;//返回的時候指針反向
	}
	else
	{
		phead = s;//記錄新的第一個結點
	}

}
void reverse1(ListNode *head)
{
	if (head == NULL)
		return;
	ListNode *p = head->next;//注意遞歸的起始條件
	ListNode *s = NULL;
	reverse2(s,p);
}
void main()
{
	List head;
	int ar[] = {12,23,34,45,56,67,78,89,90,100};
	int len = sizeof(ar)/sizeof(ar[0]);
	head= init_list();
	for (int i = 0;i<len;++i)
	{
		push_front(head,ar[i]);
	}
	print(head);
	reverse1(head);
	head->next = phead;
	print(head);

}