环形链表||（leetcode中级篇一百四十二题）

给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 null。

说明：不允许修改给定的链表。

代码如下

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB)//代码比较长其实
//只有三部分
{
	int lenA = 0,lenB = 0,gap = 0;//找俩个链表的交点，看不懂的参考笔者前期博客
	struct ListNode* a = headA;
	struct ListNode* b = headB;
	struct ListNode* meet = NULL;
	while(a != NULL && a -> next != NULL)
	{
		lenA++;
		a = a -> next;
	}
	while(b != NULL && b -> next != NULL)
	{
		lenB++;
		b = b -> next;
	}
	if(b != a)
	{
		return NULL;
	}
	else
	{
		a = headA;b = headB;
		if(lenA > lenB)
		{
			gap = lenA - lenB;
			while(gap--)
			{
				a = a -> next;
			}
		}
		else
		{
			gap = lenB - lenA;
			while(gap--)
			{
				b = b -> next;
			}
		}
		while(a && b)
		{
			if(a == b)
			{
				meet = a;
				return a;
			}
			a = a -> next;
			b = b -> next;
		}
	}
	return meet;
}
struct ListNode *hasCycle(struct ListNode *head) {//判断这个链表中有没有环
    if(head == NULL || head -> next == NULL) return false;
    struct ListNode* slow = head;
    struct ListNode* fast = head -> next;
    while(slow != fast)
    {
        if(fast == NULL || fast -> next == NULL)
        {
            return NULL;
        }
        slow = slow -> next;
        fast = fast -> next -> next;
    }
    return fast;
}
struct ListNode *detectCycle(struct ListNode *head) {//把俩个函数调用的主逻辑
    struct ListNode* h = NULL;
    struct ListNode* Union = NULL;
    struct ListNode* meet = hasCycle(head);
    if(meet == NULL)
    {
        return NULL;
    }
    else
    {
        h = meet -> next;
        meet -> next =NULL;
        Union = getIntersectionNode(head,h);
        return Union;
    }
}

这种办法就是将我们之前练习的俩个函数组合起来，接下来我们来看看方法二

struct ListNode *hasCycle(struct ListNode *head) {
    if(head == NULL || head -> next == NULL) return false;
    struct ListNode* slow = head;
    struct ListNode* fast = head -> next;
    while(slow != fast)
    {
        if(fast == NULL || fast -> next == NULL)
        {
            return NULL;
        }
        slow = slow -> next;
        fast = fast -> next -> next;
    }
    return fast;
}
struct ListNode *detectCycle(struct ListNode *head) {
    struct ListNode* meet = hasCycle(head);
    if(meet == NULL)
    {
        return NULL;
    }
    else
    {
        while(1)//一点有相交节点所以可以写成1
        {
            if(meet == head)
            {
                return meet;
            }
            meet = meet -> next;
            head = head -> next;
        }
    }
    return;    
}

ps：如果有什么问题欢迎留言与笔者讨论。