题解 | #链表中环的入口结点#

struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
        val(x), next(NULL) {
    }
};

class Solution {
public:
    ListNode* EntryNodeOfLoop(ListNode* pHead) {
        ListNode *fast = pHead, *slow = pHead;
        while(fast && fast->next){
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow){
                break; //表示可能有环
            }
        }
        //如果走到了NULL 那表示肯定没有环
        if(!fast || !fast->next){
            return nullptr;
        }
        //不然的话，把fast和slow中任意一个结点重置为pHead
        slow = pHead;
        while(slow != fast){
            slow = slow -> next;
            fast = fast -> next;
        }
        return fast;
    }
};