题目
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。
样例
给出 ⑵1->10->4->5, tail connects to node index 1，返回10
分析
上1题的进阶。
首先，利用快慢指针判断有没有环，若遇到slow == fast时，跳出循环；
然后，调剂fast=head，slow不变，此时slow与fast同步移动，直至再次相遇，即是链表中环的起始节点。
Python代码
"""
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
GitHub -- Python代码
C++代码
/**
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;
}
};
GitHub -- C++代码