判断俩单链表是否相交

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AC 代码总的来说,单链表分为两种无环单链表,有环单链表,判断单链表相交则可转化为俩无环单链表是否相交,俩有环单链表是否相交,而一条有环一条无环一定不可能相交!

class ChkIntersection {    ListNode * findLoopNode(ListNode* head,int &len){//查找入环借点,若无环返回NULL;若无环即得到无环单链表长度        ListNode *quick=NULL,*slow=head->next;        ++len;        if(!slow)            return NULL;        quick = slow->next;        ++len;        while (true) {            if(!quick||!slow)                return NULL;            if(quick == slow)                break;            slow = slow->next;            quick = quick->next;            ++len;            if(!quick)                return NULL;            quick = quick->next;            ++len;        }        quick = head;        while (quick!=slow) {            quick = quick->next;            slow = slow->next;        }        return quick;    }    ListNode *walknums(ListNode *head,int nums){//让长的链表先走nums 步        while (nums--) {            head = head->next;        }        return head;    }public:    bool chkInter(ListNode* head1, ListNode* head2, int adjust0, int adjust1) {        if(!head1||!head2)            return false;        int len1 = 0,len2 = 0;        ListNode * loop1 = findLoopNode(head1,len1), *loop2 = findLoopNode(head2,len2);        if((loop1==NULL&&loop2)||(loop1&&loop2==NULL))//一个有环 ,一个无环            return false;        else if(loop1==NULL&&loop2==NULL){//俩个都无环            if(len1>=len2)                head1 = walknums(head1,len1-len2);            else                head2 = walknums(head2,len2-len1);            while (head1&&head2) {                if(head2==head1)                    return true;                head1 = head1->next;                head2 = head2->next;            }            return false;        }else{//俩有环单链表相交            if(loop1==loop2)                return true;            else{                ListNode * p =loop1->next;                while (p!=loop1) {                    if(p==loop2)                        return true;                    p = p->next;                }                return false;            }        }    }};

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