链表操作

已知两个链表head1 和head2 各自有序 请把它们合并成一个链表依然有序

1.非递归

Node* Merge(Node* head1, Node* head2)
{
	Node* head = NULL
        Node* head_temp = NULL;
	if(head1 == NULL) return head2;
	if(head2 == NULL) return head1;
	Node* p = head1;
	Node* q = head2;
	
	if(p->value < q->value)
	{
		head_temp= p;
		p = p->next;
	}
	else
	{
		head_temp = q;
		q = q->next;
	}
	head = head_temp;
	while(NULL != p && NULL != q)
	{
		if(p->value < q->value)
		{
			head_temp->next = p;
			head_temp = p;
			p = p->next;
		}
		else
		{
			head_temp->next = q;
			head_temp = q;
			q = q->next;
		}
	}
	while(p)
	{
	         head_temp->next = p;
                  head_temp = p;
		p = p->next;
	}
	while(q)
	{
		head_temp->next = q;
		head_temp = q;
		q = q->next;
	}
	return head;
}

2. 递归

struct Node
{
 Node* next;
 int value;
};
Node* Merge(Node* head1, Node* head2)
{
	Node* res;
	if(head1 == NULL) return head2;
	if(head2 == NULL) return head1;
	if(head1->value < head2->value)
	{
		res = head1;
		res->next = Merge(head1->next, head2);
	}
	else
	{
		res = head2;
		res->next = Merge(head1, head2->next);
	}
	return res;
}

单链表的逆置

LNode*  ReverseList(LNode* head)
{
	if (head == NULL)
		return NULL;
	LNode* pre = head;
	LNode* cur = head->next;
	LNode* nxt = cur;
	while (nxt!=NULL)
	{
		nxt = cur->next;
		cur->next = pre;
		pre = cur;
		cur =nxt;
	}
      return pre;
}