141.142.环形链表

思路：（1）设置两个快慢指针fast和slow。

（2）fast一次走两步，slow每次走一步。

（3）如果有环，那么fast一定会在环中和slow相遇。如果没环，则不会。

代码为

bool hasCycle(struct ListNode *head) {
   struct ListNode *slow=head;
   struct ListNode *fast=head;
   while(fast&&fast->next)
   {
      slow=slow->next;
      fast=fast->next->next;
      if(slow==fast)
      {
          return true;
      }
   } 
return false;
}

证明：（1）快指针是如何追上慢指针的？

           （2）如果慢指针一次走一步，fast指针走两步，那么fast走三步，走四步行不行？ 

思路：方法一：利用链表相交的原理，fast和slow一定会在环中相遇，相遇点设为meet。

 然后从meet这里断开，这下问题就转变为求两个链表相交，求相交节点。

 代码为

//求两个链表相交的交点
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode *curA=headA;
    struct ListNode *curB=headB;
    int lengthA=0,lengthB=0;
    while(curA)
    {
        lengthA++;
        curA=curA->next;
    }
    while(curB)
    {
       lengthB++;
        curB=curB->next;
    }
    int gap=abs(lengthA-lengthB);
    struct ListNode * less=headA;
    struct ListNode * great=headB;
  
        if(lengthA>lengthB)
        {
           great=headA;
        less=headB;
        } 
     
        while(gap--)
        {
      great=great->next;
        }
        while(great)
        {
            if(great==less)
            return great;
            else{
                great=great->next;
                less=less->next;
            }
           
        }
 return NULL;
}
struct ListNode *detectCycle(struct ListNode *head) {
   struct ListNode *slow=head;
   struct ListNode *fast=head;

   while(fast&&fast->next)
   {
      slow=slow->next;
      fast=fast->next->next;
      if(slow==fast)
      {
          break;
      }
   }
    if(fast==NULL||fast->next==NULL)
    return NULL;
       struct ListNode *list2=fast->next;
          fast->next=NULL;
      struct ListNode *meet= getIntersectionNode(head,list2);
return  meet;
}

方法二：

代码为

struct ListNode *detectCycle(struct ListNode *head) {
   struct ListNode *slow=head;
   struct ListNode *fast=head;

   while(fast&&fast->next)
   {
      slow=slow->next;
      fast=fast->next->next;
      if(slow==fast)
      {
          break;
      }
   } 
   if(fast==NULL||fast->next==NULL)
   return NULL;   
   struct ListNode *meet=fast;
   while(meet!=head)
   {
       meet=meet->next;
       head=head->next;
   }
   return meet;

}