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) {
        if (!pHead || !pHead -> next) return nullptr;
        ListNode* slow = pHead, *fast = pHead;

        while (fast) {
            slow = slow -> next;
            
            if (!fast -> next) return nullptr;
            fast = fast -> next;
            if (!fast -> next) return nullptr;
            fast = fast -> next;

            if (slow == fast) break;
        }
        // fast = fast -> next;

        slow = pHead;
        while (true) {
            if (slow == fast) return slow;
            slow = slow -> next;
            fast = fast -> next;
        }

        return nullptr;
    }
};

注意下，fast和slow的其实位置如果不相同，则找到位置之后，fast需要后进一位呢