两种方法求单链表逆序

1 递归,非常easy

代码:

#include<iostream>  

using namespace std;  

typedef struct node{
	int data;
	struct node * pNext;
}Node ,*pNode;

void createNode(pNode & pHead){
	int temp;
	scanf("%d",&temp);
	pNode p,q;
	bool isFirst = true;
	while(temp){
		if(isFirst){
			p=q=pHead=(pNode)malloc(sizeof(Node));
			pHead->data = temp;
			isFirst = false;
		}else{
			q = new Node();
			q->data = temp;
			q->pNext=NULL;
			p->pNext = q;
			p =q;
		}
		scanf("%d",&temp);
	}

}
void print(pNode pHead){
	if(pHead){
		cout<<pHead->data<<" ";
		print(pHead->pNext);
	}else
	return;		
}

void reverse(pNode pcur,pNode & pHead){
	if((NULL == pcur) ||( NULL == pcur->pNext)){
			pHead = pcur;
			return;
	}else{
		pNode pnext = pcur->pNext;
		reverse(pnext,pHead);
		pnext->pNext = pcur;
		pcur->pNext = NULL;
	}
}
int main()  
{  
	pNode pHead = NULL;
	createNode(pHead);
	cout<<"原链表:"<<endl;
	print(pHead);
	cout<<"翻转:"<<endl;
	reverse(pHead,pHead);
	print(pHead);
    return 0;  
}  

执行结果:


2 不是递归,循环:

#include<iostream>  

using namespace std;  

typedef struct node{
	int data;
	struct node * pNext;
}Node ,*pNode;

void createNode(pNode & pHead){
	int temp;
	scanf("%d",&temp);
	pNode p,q;
	bool isFirst = true;
	while(temp){
		if(isFirst){
			p=q=pHead=(pNode)malloc(sizeof(Node));
			pHead->data = temp;
			isFirst = false;
		}else{
			q = new Node();
			q->data = temp;
			q->pNext=NULL;
			p->pNext = q;
			p =q;
		}
		scanf("%d",&temp);
	}

}
void print(pNode pHead){
	if(pHead){
		cout<<pHead->data<<" ";
		print(pHead->pNext);
	}else
	return;		
}

pNode reverse(pNode & pHead){
	if(NULL == pHead || NULL == pHead->pNext){
		return pHead;
	}
	pNode p1,p2,p3;
	p1=pHead;
	p2=pHead->pNext;
	while(p2){
		p3=p2->pNext;
		p2->pNext = p1;
		p1 = p2;
		p2 = p3;
	}
	pHead->pNext = NULL; /*设置链表尾*/
	pHead = p1;  /*调整链表头*/
	return pHead;
}
int main()  
{  
	pNode pHead = NULL;
	createNode(pHead);
	cout<<"原链表:"<<endl;
	print(pHead);
	cout<<"翻转:"<<endl;
	reverse(pHead);
	print(pHead);
    return 0;  
}  

执行结果:



原文地址:https://www.cnblogs.com/gcczhongduan/p/4057968.html