链表逆置

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同学提到了这个问题，自己动手编了一下。

有两种方法：递归和非递归。

递归中又有两种：带返回值的和不带返回值的。

代码如下：

/****    链表的逆置      *****//*struct node {int a;node* next;node(): a(0), next(NULL) {}};void Creat_list(node* &head, int n){node* p = new node;p->a = n;if(!head)head->next = p;else {p->next = head->next;head->next = p;}}void show(node* &head){node* p = head->next;while(p) {cout << p->a << "  ";p = p->next;}cout << endl;}//***    非递归实现   void reverse(node* &head)            //带头结点的链表反转{node* pre = head->next;node* cur = pre->next;node* nex = NULL;if(!pre || !cur)return;while(cur) {nex = cur->next;cur->next = pre;pre = cur;cur = nex;}head->next->next = NULL;head->next = pre;}//递归实现（带返回值） node* reverse(node* &head, node* cur){if(!cur || !cur->next) {head->next = cur;return cur;}else {node* tmp = reverse(head, cur->next);tmp->next = cur;cur->next = NULL;return cur;}}//    递归实现（ 不 带返回值）   void reverse1(node* &head, node* cur){if(!cur || !cur->next) {head->next = cur;}else {reverse(head, cur->next);cur->next->next = cur;cur->next = NULL;}}int main(){node* head = new node;Creat_list(head, 1);Creat_list(head, 2);Creat_list(head, 3);Creat_list(head, 4);Creat_list(head, 5);Creat_list(head, 6);show(head);cout << endl;//reverse(head);               //非递归node* p = head->next;//node* a = reverse(head, p);   // 递归带返回值   （其实不用带返回值，带返回值的较复杂）reverse1(head, p);show(head);getchar();return 0;}

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