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* slow = pHead;
        ListNode* fast = pHead;
        while(fast!=nullptr)
        {
            fast = fast->next;
            if(fast==nullptr)  return nullptr;
            slow = slow->next;
            fast = fast->next;
            if(fast==nullptr)  return nullptr;
            if(fast==slow)  break;
        }
        fast = pHead;
        while(fast!=slow)
        {
            fast = fast->next;
            slow = slow->next;
        }
        return slow;
    }
};

头结点到入口节点距离与 快慢指针相遇处到入口节点的距离相等

第一次相遇后, 快指针置于phead,和慢指针一样每次只走一步, 再次相遇即是入口节点