142.环形链表 II（LeetCode）

想法一 

 由环形链表题中，沿用快慢指针思想，再结合以下结论

https://blog.csdn.net/2301_79188764/article/details/134299433 

 

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

    while (fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;

        if (slow == fast)
        {
            struct ListNode *meet = slow;

            while (head != meet)
            {
                head = head->next;
                meet = meet->next;
            }

            return meet;
        }
    }

    return NULL;
}

结论分析和推导 

 

 

 

 

好像分析的很对，是吧？但是，很遗憾的告诉你，这个推论是错的！ 

 

 

这样，推论才是对的，前面只是特殊情况，刚好符合结论而已  

想法二 

将链表的环切断，转换为链表相交问题

https://blog.csdn.net/2301_79188764/article/details/134298439 

 

思路比较简单，没有用数学结论，但是代码比较长 

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB)
{
    struct ListNode *tail1 = headA, *tail2 = headB;
    int len1 = 1, len2 = 1;
    
    while (tail1->next)
    {
        tail1 = tail1->next;
        len1++;
    }

    while (tail2->next)
    {
        tail2 = tail2->next;
        len2++;
    }

    if (tail1 != tail2)
    {
        return NULL;
    }

    int gap = abs(len1 - len2);
    struct ListNode *longlist = headA, *shortlist = headB;

    if (len1 < len2)
    {
        longlist = headB;
        shortlist = headA;
    }

    while (gap--)
    {
        longlist = longlist->next;
    }

    while (longlist != shortlist)
    {
        longlist = longlist->next;
        shortlist = shortlist->next;
    }

    return longlist;
}

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

    while (fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;

        if (slow == fast)
        {
            struct ListNode *meetNext = slow->next;
            slow->next = NULL;
            return getIntersectionNode(head, meetNext);
        }
    }

    return NULL;
}