leetcode链表--6linked-list-cycle-ii(有环单链表环的入口结点)

Posted 2020-09-17 qqky

题目描述

 

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

Follow up:

Can you solve it without using extra space?

 

解题思路：

1、确定环中结点的数目（快慢指针确定环中结点，然后遍历到下一次到该节点确定数目n）

2、一个指针先走n步，一个指针指向头结点

3、然后两个指针一起走

4、相遇点为入口点

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     //1、确定环中结点数目n
12     //2、两个指针  1个先走n步，然后两个一起走
13     //3、相遇点即为入口点
14     ListNode *detectCycle(ListNode *head) {
15         if(head == NULL)
16             return NULL;
17         ListNode *node = MettingNode(head);
18         if(node == NULL)//无环
19         {
20             return NULL;
21         }
22         int nodeNum = 1;
23         ListNode *pNode = node;
24         while(pNode->next != node)
25         {
26             pNode =pNode->next;
27             nodeNum++;
28         }
29         pNode = head;//pNode指向头结点
30         for(int i=0;i<nodeNum;i++)
31         {
32             pNode = pNode->next;
33         }
34         ListNode *pNode2  = head;
35         while(pNode2 != pNode)//两结点不相遇
36         {
37             pNode = pNode->next;
38             pNode2 = pNode2->next;
39         }
40         return pNode;
41     }
42     //快慢指针找环中相遇结点
43     //找到相遇节点就可确定环中结点数目
44     ListNode *MettingNode(ListNode *head)
45     {
46         if(head == NULL)
47             return NULL;
48         ListNode *slow = head;
49         ListNode *fast = head;
50         while(fast != NULL && fast->next != NULL)
51         {
52             slow = slow->next;
53             fast = fast->next->next;
54             if(slow == fast)
55                 return slow;
56         }
57         return NULL;
58     }
59 };