""" Definition of ListNode class ListNode(object): def __init__(self, val, next=None): self.val = val self.next = next """ class Solution: """ @param head: The first node of the linked list. @return: The node where the cycle begins. if there is no cycle, return null """ def detectCycle(self, head): # write your code here if head == None or head.next == None: return None slow = head fast = head while fast != None and fast.next != None: slow = slow.next fast = fast.next.next if slow == fast: break if fast != None and fast == slow: fast = head while fast != slow: slow = slow.next fast = fast.next return fast return None

/** 103 带环链表 II 给定1个链表，如果链表中存在环，则返回到链表中环的起始节点的值，如果没有环，返回null。 您在真实的面试中是不是遇到过这个题？ Yes 样例 给出 ⑵1->10->4->5, tail connects to node index 1，返回10 */ /** * Definition of ListNode * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: The first node of linked list. * @return: The node where the cycle begins. * if there is no cycle, return null */ ListNode *detectCycle(ListNode *head) { // write your code here if(head == NULL || head->next ==NULL) { return NULL; }//if ListNode *slow = head, *fast = head; while(fast && fast->next) { slow = slow->next; fast = fast->next->next; if(slow == fast) { break; }//if }//while if(fast && fast == slow) { fast = head; while(fast != slow) { fast = fast->next; slow = slow->next; }//while return fast; }//if return NULL; } };