Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:

Can you solve it without using extra space?

解题思路：

（1）用快慢指针判断链表是否有环，快指针每次走2步，慢指针每次走1步，若两者相遇则有环，若遇到NULL则没环

（2）如果有环，计算换的长度count

（3）回到链表头，用先后指针寻找环入口结点，先指针先走count步，然后先后指针以相同的速度移动，当两者相遇时就是环的入口结点

struct ListNode *detectCycle(struct ListNode *head) {

bool flags = false;
int count = 1;
struct ListNode *slowpNode = head;
struct ListNode *fastpNode = head;

//寻找是否有环
while ((fastpNode != NULL) && (fastpNode->next != NULL))
{
slowpNode = slowpNode->next;
fastpNode = fastpNode->next->next;
if (slowpNode == fastpNode)
{
flags = true;
break;
}
}
if (!flags)
return NULL;

//判断环的长度
fastpNode = fastpNode->next;
while (slowpNode != fastpNode)
{
count++;
fastpNode = fastpNode->next;
}

//寻找环的入口结点
slowpNode = head;
fastpNode = head;
while (count--)
{
fastpNode = fastpNode->next;
}
while (slowpNode != fastpNode)
{
slowpNode = slowpNode->next;
fastpNode = fastpNode->next;
}

return slowpNode;

}