链表中的环

第一种：快慢指针解决环问题

快指针一次移动两次，慢指针一次移动一次，判断条件为快慢指针相遇，如果快指针为空（表示没环）如果相遇则有环，则快指针指向初试，然后快慢指针每次移动一次，直到相遇即环入口。

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
        val(x), next(NULL) {
    }
};
*/
class Solution {
public:
    ListNode* EntryNodeOfLoop(ListNode* pHead) {
        if(pHead == NULL || pHead->next == NULL)return NULL;
        ListNode  *fast  = pHead->next->next, *slow = pHead->next;
        while(fast != slow){
            slow = slow->next;
            if(fast == NULL || fast->next == NULL)return NULL;
            fast = fast->next->next;
        }
        fast = pHead;
        while(fast != slow){
            fast = fast->next;
            slow = slow->next;
        }
        return fast;
    }
};

方法二：hashset保存，如果有重复即环入口

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
        val(x), next(NULL) {
    }
};
*/
class Solution {
public:
    ListNode* EntryNodeOfLoop(ListNode* pHead) {
        unordered_set<ListNode*> s;
        while(pHead){
        if(s.count(pHead))return pHead;
        s.insert(pHead);
         pHead = pHead->next;
        }
        return NULL;
    }
};