4.23 链表环操作

学学别人写一点总结也不错。睡了一个上午，下午还看了2场dota比赛，一天没学啥啊，明天要改变。

今天看了一下链表的东西，看到几个有用的东西。

1.如何判断链表是否有环。

It's easy to determine whether the link has a circle. Suppose there are two points P and Q. P and Q start from the head point. Every time Q move one step ,P move two steps. If circle exits, P and Q will meet.

bool hasLoop(node* head){
    if(NULL==head)
        return true;
    node* p=head;
    node* q=head;
    while(p!=NULL&&p->next!=NULL){
        p=p->next->next;
        q=q->next;
        if(p==q)
            return true;
    }
    return false;
}

 

2.环的长度。

if you are sure the link has a circle, you can calculate the length.  Assume point Q and P .Useing the method above ,P and Q will meet. When they meet first，allow Q move step by step . When p meet itself again,the steps is the circle length.

 

int looplen(node* head){
    node *p=head;
    node *q=head;
    bool meet=false;
    int len=0;
    while(p&&p->next){
        if(!meet){
            p=p->next->next;
            q=q->next;
            if(p==q){
                meet=true;
            }
        }
        else{
            len++;
            q=q->next;
            if(q==p)
                break;
        }
    }
    return len;
}

 

3.环的入口。

if you know the length of the circle is n .  Assume Point P and Q again.  Allow p move n steps fisrt. Then P and Q move step by step, when P meet Q , the p is the start point of the circle.

node* LoopEntrance(node* head){
    node*p =head,*q=head;
    int len=circlelen(head);
    while(len--)
        p=p->next;
    while(p!=q){
        p=p->next;
        q=q->next;
    }
    return p;
}