LintCode（103）带环链表 II

题目

给定一个链表，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否遇到过这个题？

Yes

样例

给出 -21->10->4->5, tail connects to node index 1，返回10

分析

上一题的进阶。
首先，利用快慢指针判断有无环，若遇到slow == fast时，跳出循环；
然后，调整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

给定一个链表，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否遇到过这个题？ Yes
样例
给出 -21->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