Android 笔试---C链表

（1）实现链表的逆置:可以参考http://blog.csdn.net/heyabo/article/details/7610732(有示意图)

node *inverselinklist(node *head)

{
        node *p1,*p2,*p3;
        if(NULL==head||NULL==head->next) {
               return head;
        }
        p1=head;
        p2=p1->next;
        while (p2) {
               p3=p2->next;
               p2->next=p1;
               p1=p2;
               p2=p3;
        }
        head->next=NULL;
        head=p1;
        return head;
}

（2）用普通算法实现两个有序链表的合并

  

node *mergelinklist(node *head1,node *head2)
{
    node *p1,*p2,*temp=NULL,*cur=NULL,*head=NULL;
    p1=head1;
    p2=head2;
    while (p1!=NULL&&p2!=NULL) {

        if (p1->value<p2->value) {
            temp=p1;
            p1=p1->next;
        }
        else
        {
            temp=p2;
            p2=p2->next;
        }
        if (NULL==head) {
            cur=temp;
            head=cur;
        }
        else
        {
            cur->next=temp;
            cur=temp;
        }

    }
    temp=(p1?p1:p2);
    while (temp!=NULL) {
            cur->next=temp;
            cur=cur->next;
            temp=temp->next;
    }
    cur->next=NULL;
    return head;
}

 

（3）用递归算法实现两个有序列表的合并

node *mergeRecursion(node *head1,node *head2)
{
        if(NULL==head1)
        {
               return head2;
        }
        if(NULL==head2)
        {
               return head1;
        }
        node *head=NULL;
        if (head1->value<=head2->value) {
              head=head1;
              head->next=mergeRecursion(head1->next,head2);
        }
        else
        {
               head=head2;
               head->next=mergeRecursion(head1,head2->next);
        }
        return head;
}

（4）判断单链表中是否存在环

bool IsExitsLoop(slist *head)
{
    slist *slow = head, *fast = head;
    while ( fast && fast->next )
    {
        slow = slow->next;
        fast = fast->next->next;
        if ( slow == fast ) break;
    }
    return !(fast == NULL || fast->next == NULL);
}