单链表逆序 也叫反转

I have browsed many codes about the list reversed, most of which are hard to understand and complicated;

to reverse a single list, two methods can be required:

1; create a another list from the old list reversed, that is to say the new nodes are coming from the tail of the old list.

2; reverse the pointer; we need only to reverse the pointer from next to prior;

so we found the last method are better than first , because it is terse and without allocating new memory;

for example, original order is following:

prior      cur         next

○->○->○

reverse

○<-○<-○

ok, principle is easy to understand, implement it

#includeusing namespace std;struct Node{int data;struct Node *next;};struct Node *createList(const int & aiNum){struct Node *head = NULL;struct Node *newPtr = NULL;struct Node *cur;int i = 0;for (; i < aiNum; i++ ){if (!head){head = (struct Node*)malloc(sizeof(Node));head->data = i;head->next = NULL;cur = head;}else{newPtr = (struct Node*)malloc(sizeof(Node));newPtr->data = i;newPtr->next = NULL;cur->next = newPtr;    cur = cur->next;}}cur->next = NULL;return head;}void printNode(struct Node * head){cout<<"print node"<<endl;struct Node * lptr = head;while (lptr){cout<<lptr->data<<endl;lptr = lptr->next;}}void destroyList(struct Node * head){cout<<"destroy list"<<endl;struct Node *cur = head;int i = 0;while (head){cur = head;cout<<cur->data<<endl;head = head->next;free(cur);}}Node* reverseList(struct Node * head){Node *prior = head;Node * cur = prior->next;Node * next = NULL;while (cur->next)//prior->cur->next{next = cur->next;cur->next = prior;//prior<-cur->nextprior = cur;      //      prior cur  ;move to the next node for next loopcur = next;}head->next = NULL;   //set the tail as NULLcur->next = prior;  //the last loop , since the cur->next is null ,                   //the pointer is still from head to cur,so we must reverse it additionalprintNode(cur);return cur;}void main(){struct Node *head = createList(5);printNode(head);struct Node *rhead = reverseList(head);destroyList(rhead);}