Linked List Cycle II

一、问题描述

Given a linked list, return the node where the cycle begins. If there is no cycle, return 

null

.

Note: Do not modify the linked list.
二、思路

判断单链表是否有环，并且找到入口请见我另一篇博客：判断单链表是否存在环以及求出环的入口

三、代码

class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *fast = head;
ListNode *slow = head;
bool flag = false;
while(fast && fast -> next && fast -> next -> next){
fast = fast -> next -> next;
slow =slow -> next;
if(fast == slow){
flag = true;
break;
}
}
if(!flag)
return NULL;
slow = head;
while(slow != fast){
slow = slow -> next;
fast = fast -> next;
}
return fast;
}
};